2.8 Exercise Problems and Solutions

Q1. (i) Calculate the number of electrons that will together weigh one gram.
(ii) Calculate the mass and charge of one mole of electrons.

Answer:(i) Mass of an electron \(=9.1 \times 10^{-28} \mathrm{~g}\)
\(9.1 \times 10^{-28} \mathrm{~g}\) is the mass of \(=1\) electron
\(1.0 \mathrm{~g}\) is the mass of \(=\frac{1}{9.1 \times 10^{-28}}=\mathbf{1 . 0 9 9} \times \mathbf{1 0 ^ { 2 7 }}\) electrons

(ii) One mole of electrons \(=6.022 \times 10^{23}\) electrons
Mass of 1 electron \(=9.1 \times 10^{-31} \mathrm{~kg}\)
Mass of \(6.022 \times 10^{23}\) electrons \(=(9.1 \times 10.31 \mathrm{~kg}) \times\left(6.022 \times 10^{23}\right)=5.48 \times 10^{-7} \mathrm{~kg}\)
Charge on one electron \(=1.602 \times 10^{-19}\) coulomb
Charge on one mole electrons \(=1.602 \times 10^{-19} \times 6.022 \times 10^{23}=9.65 \times 10^4\) coulombs

Q2. (i) Calculate the total number of electrons present in one mole of methane.
(ii) Find (a) the total number and (b) the total mass of neutrons in \(7 \mathrm{mg}\) of \({ }^{14} \mathrm{C}\). (Assume that mass of a neutron \(=1.675 \times 10^{-27} \mathrm{~kg}\) ).
(iii) Find (a) the total number and (b) the total mass of protons in \(34 \mathrm{mg}\) of \(\mathrm{NH}_3\) at STP.
Will the answer change if the temperature and pressure are changed?

Answer: (i) One mole of methane \(\left(\mathrm{CH}_4\right)\) has molecules \(=6.022 \times 10^{23}\)
No. of electrons present in one molecule of \(\mathrm{CH}_4=6+4=10\)
No. of electrons present in \(6.022 \times 10^{23}\) molecules of \(\mathrm{CH}_4=6.022 \times 10^{23} \times 10\) \(=6.022 \times 10^{24}\) electrons
(ii) Step I. Calculation of the total number of carbon atoms
Gram atomic mass of carbon \((\mathrm{C}-14)=14 \mathrm{~g}=14 \times 10^3 \mathrm{mg}\)
\(14 \times 10^3 \mathrm{mg}\) of carbon \((\mathrm{C}-14)\) have atoms \(=6.022 \times 10^{23}\)
\(7 \mathrm{mg}\) of carbon \((\mathrm{C}-14)\) have atoms \(=\frac{6.022 \times 10^{23}}{\left(14 \times 10^3 \mathrm{mg}\right)} \times(7 \mathrm{mg})=3.011 \times 10^{20}\) atoms.
Step II. Calculation of the total number and total mass of neutrons
No. of neutrons present in one atom (C-14) of carbon \(=14-6=8\)
No. of neutrons present in \(3.011 \times 10^{20}\) atoms \((\mathrm{C}-14)\) of carbon \(=3.011 \times 10^{20} \times 8\) \(=2.408 \times 10^{21}\) neutrons
Mass of one neutron \(=1.675 \times 10^{-27} \mathrm{~kg}\)
Mass of \(2.408 \times 10^{21}\) neutrons \(=\left(1.675 \times 10^{-27} \mathrm{~kg}\right) \times 2.408 \times 10^{21}\) \(=4.033 \times 10^{-6} \mathrm{~kg}\).
(iii) Step I. Calculation of total number of \(\mathrm{NH}_3\) molecules
Gram molecular mass of ammonia \(\left(\mathrm{NH}_3\right)=17 \mathrm{~g}=17 \times 10^3 \mathrm{mg}\) \(17 \times 10^3 \mathrm{mg}\) of \(\mathrm{NH}_3\) have molecules \(=6.022 \times 10^{23}\)
\(
\begin{aligned}
34 \mathrm{mg} \text { of } \mathrm{NH}_3 \text { have molecules } & =\frac{6.022 \times 10^{23}}{\left(17 \times 10^3 \mathrm{mg}\right)} \times(34 \mathrm{mg}) \\
& =\mathbf{1 . 2 0 4 4} \times 10^{20} \text { molecules. }
\end{aligned}
\)
Step II. Calculation of the total number and mass of protons
No. of protons present in one molecule of \(\mathrm{NH}_3=7+3=10\).
No. of protons present in \(12.044 \times 10^{20}\) molecules of \(\mathrm{NH}_3=12.044 \times 10^{20} \times 10\)
\(=1.2044 \times 10^{22}\) protons
Mass of one proton \(=1.67 \times 10^{-27} \mathrm{~kg}\)
Mass of \(1.2044 \times 10^{22}\) protons \(=\left(1.67 \times 10^{-27} \mathrm{~kg}\right) \times 1.2044 \times 10^{22}\) \(=2.01 \times 10^{-5} \mathrm{~kg}\).
No, the answer will not change upon changing the temperature and pressure because only the number of protons and mass of protons are involved.

Q3. How many neutrons and protons are there in the following nuclei?
\(
{ }_6^{13} \mathrm{C},{ }_8^{16} \mathrm{O},{ }_{12}^{24} \mathrm{Mg},{ }_{26}^{56} \mathrm{Fe},{ }_{38}^{88} \mathrm{Sr}
\)

Answer: (i) \(\quad{ }_6^{13} \mathrm{C}\); Atomic no. (Z) \(=6\)
No. of protons \((p)=6\)
Mass no. (A) \(=13\)
No. of neutrons \((n)=13-6=7\)
(ii) \(\quad{ }_8^{16} \mathrm{O}\); Atomic no, \((\mathrm{Z})=8\)
Mass no. \((\mathrm{A})=16\)
No. of protons \((p)=\mathbf{8}\)
No. of neutrons \((n)=16-8=\mathbf{8}\)
(iii) \({ }_{12}^{24} \mathrm{Mg}\); Atomic no. (Z) \(=12\)
Mass no. \((\mathrm{A})=24\)
No. of protons \((p)=12\)
No. of neutrons \((n)=24-12=12\)
(iv) \({ }_{26}^{56} \mathrm{Fe} ;\) Atomic no (Z) \(=26\)
Mass no. \((A)=56\)
No. of protons \((p)=\mathbf{2 6}\)
No. of neutrons \((n)=56-26=30\).
(v) \(\quad{ }_{38}^{88} \mathrm{Sr} ;\) Atomic no \((\mathrm{Z})=38\)
Mass no. (A) \(=88\)
No. of protons \((p)=\mathbf{3 8}\)
No. of neutrons \((n)=\mathbf{5 0}\).

Q4. Write the complete symbol for the atom with the given atomic number \((\mathrm{Z})\) and atomic mass (A)
(i) \(Z=17, A=35\).
(ii) \(Z=92, A=233\).
(iii) \(\mathrm{Z}=4, \mathrm{~A}=9\).

Answer: (i)
\(
\begin{gathered}
{ }_{17}^{35} \mathrm{X} \\
(\mathrm{X}=\mathrm{Cl})
\end{gathered}
\)
(ii)
\(
\begin{gathered}
{ }_{92}^{233} \mathrm{X} \\
(\mathrm{X}=\mathrm{U})
\end{gathered}
\)
(iii)
\(
\begin{gathered}
{ }_4^9 \mathrm{X} \\
(\mathrm{X}=\mathrm{Be})
\end{gathered}
\)

Q5. Yellow light emitted from a sodium lamp has a wavelength \((\lambda)\) of \(580 \mathrm{~nm}\). Calculate the frequency \((f)\) and wavenumber \((\tilde{\nu})\) of the yellow light.

Answer: Step I. Calculation of frequency of yellow light
\(
\begin{aligned}
& \text { We know that } \quad \begin{aligned}
f & =\frac{c}{\lambda} \\
c & =3 \times 10^8 \mathrm{~m} \mathrm{~s}^{-1} ; \lambda=580 \mathrm{~nm}=580 \times 10^{-9} \mathrm{~m} \\
\therefore \quad f & =\frac{\left(3 \times 10^8 \mathrm{~m} \mathrm{~s}^{-1}\right)}{\left(580 \times 10^{-9} \mathrm{~m}\right)}=\mathbf{5} \cdot 17 \times 10^{14} \mathrm{~s}^{-1}
\end{aligned}
\end{aligned}
\)
Step II. Calculation of wave number of yellow light
Wave number \((\tilde{\nu})=\frac{1}{\lambda}=\frac{1}{\left(580 \times 10^{-9} \mathrm{~m}\right)}=\mathbf{1 . 7 2 4} \times 10^6 \mathrm{~m}^{-1}\).

Q6. Find the energy of each of the photons which
(i) correspond to light of frequency \(3 \times 10^{15} \mathrm{~Hz}\).
(ii) have wavelength of 0.5 Å.

Answer: (i) Energy of photon \((E)=h f\)
\(
h=6.626 \times 10^{-34} \mathrm{~J} \mathrm{~s} ; {f}=3 \times 10^{15} \mathrm{~Hz}=3 \times 10^{15} \mathrm{~s}^{-1}
\)
\(
\therefore E=\left(6.626 \times 10^{-34} \mathrm{~J} \mathrm{~s}\right) \times\left(3 \times 10^{15} \mathrm{~s}^{-1}\right)=1.986 \times 10^{18} \mathrm{~J}
\)
(ii)
\(
\text { Energy of photon }(\mathrm{E})={hf}=\frac{h c}{\lambda}
\)
\(
\begin{aligned}
& {h}=6.626 \times 1034 \mathrm{~J} \mathrm{~s} ; \mathrm{c}=3 \times 10^8 \mathrm{~m} \mathrm{~s}^{-1} \\
& \lambda=0.50 Å =0.5 \times 10^{-10} \mathrm{~m} .
\end{aligned}
\)
\(
\mathrm{E}=\frac{\left(6.626 \times 10^{-34} \mathrm{~J} \mathrm{~s}\right) \times\left(3 \times 10^8 \mathrm{~ms}^{-1}\right)}{\left(0.5 \times 10^{-10} \mathrm{~m}\right)}=\mathbf{3 . 9 8} \times 10^{-15} \mathrm{~J} .
\)

Q7. Calculate the wavelength, frequency and wavenumber of a light wave whose period is \(2.0 \times 10^{-10} \mathrm{~s}\).

Answer:

\(
\text { Frequency }(f)=\frac{1}{\text { Period }}=\frac{1}{\left(2 \cdot 0 \times 10^{-10} \mathrm{~s}\right)}=\mathbf{5 . 0} \times \mathbf{1 0}^{\mathbf{9}} \mathrm{s}^{-1}
\)
\(
\text { Wavelength }(\lambda)=\frac{c}{f}=\frac{\left(3 \times 10^8 \mathrm{~ms}^{-1}\right)}{\left(5 \times 10^9 \mathrm{~s}^{-1}\right)} \quad=6.0 \times 10^{-2} \mathrm{~m}
\)
\(
\text { Wave number }(\tilde{\nu})=\frac{1}{\lambda}=\frac{1}{\left(6.0 \times 10^{-2} \mathrm{~m}\right)} \quad=\mathbf{1 6 . 6 6} \mathrm{m}^{-1} \text {. }
\)

Q8. What is the number of photons of light with a wavelength of \(4000 \mathrm{pm}\) that provide \(1 \mathrm{~J}\) of energy?

Answer: Energy of photon \((E)=\frac{h c}{\lambda}\)
\(
{h}=6.626 \times 10^{-34} \mathrm{Js}, \mathrm{c}=3 \times 10^8 \mathrm{~m} \mathrm{~s}^{-1}, \lambda=4000 \mathrm{pm}=4000 \times 10^{-12}=4 \times 10^{-9} \mathrm{~m}
\)
Energy of photon
\(
(E)=\frac{\left(6.626 \times 10^{-34} \mathrm{Js}\right) \times\left(3 \times 10^8 \mathrm{~m} \mathrm{~s}^{-1}\right)}{\left(4 \times 10^{-9} \mathrm{~m}\right)}=4.969 \times 10^{-17} \mathrm{~J}
\)
\(
\text { Now, } 4.965 \times 10^{-17} \mathrm{~J} \text { is the energy of photon }=1
\)
\(
\therefore \quad 1 \mathrm{~J} \text { is the energy of photons }=\frac{1}{4.969 \times 10^{-17}}=\mathbf{2 . 0 1 2} \times \mathbf{1 0 ^ { 1 6 }} \text { photons. }
\)

Q9. A photon of wavelength \(4 \times 10^{-7} \mathrm{~m}\) strikes on the metal surface, the work function of the metal being \(2.13 \mathrm{eV}\). Calculate (i) the energy of the photon (eV), (ii) the kinetic energy of the emission, and (iii) the velocity of the photoelectron \(\left(1 \mathrm{eV}=1.6020 \times 10^{-19} \mathrm{~J}\right)\).

Answer: (i) The energy of the photon.
\(
\text { Energy }(\mathrm{E})=\frac{h c}{\lambda}=\frac{\left(6.626 \times 10^{-34} \mathrm{Js}\right) \times\left(3 \times 10^8 \mathrm{~m} \mathrm{~s}^{-1}\right)}{\left(4 \times 10^{-7} \mathrm{~m}\right)}=4.97 \times 10^{-19} \mathrm{~J}
\)
\(
=\frac{(1 \mathrm{eV})}{\left(1.602 \times 10^{-19} \mathrm{~J}\right)} \times\left(4.97 \times 10^{-19} \mathrm{~J}\right)=\mathbf{3 . 1} \mathrm{eV}
\)
(ii) Kinetic energy of emission
The kinetic energy of emission \(=\mathrm{E}\) – work function (i.e. kinetic energy of emitted electrons)
\(
=(3.1-2.13)=0.97 e \mathbf{V}
\)
(iii) Velocity of photoelectron
\(
\mathrm{KE} \text { of emission }=\frac{1}{2} m v^2=0.97 e \mathrm{~V}
\)
\(
=0.97 \times 1.602 \times 10^{-19} \mathrm{~J}=0.97 \times 1.602 \times 10^{-19} \mathrm{~kg} \mathrm{~m}^2 \mathrm{~s}^{-2}
\)
\(
v^2=\frac{2 \times 0.97 \times 1.602 \times 10^{-19}\left(\mathrm{~kg} \mathrm{~m}^2 \mathrm{~s}^{-2}\right)}{\left(9.1 \times 10^{-31} \mathrm{~kg}\right)}=0.34 \times 10^{12} \mathrm{~m}^2 \mathrm{~s}^{-2}
\)
\(
v=\left(0.34 \times 10^{12} \mathrm{~m}^2 \mathrm{~s}^{-2}\right)^{1 / 2}=0.583 \times 10^6 \mathrm{~m} \mathrm{~s}^{-1}=\mathbf{5 . 8 3} \times 10^5 \mathrm{~m} \mathrm{~s}^{-1} .
\)

Q10. Electromagnetic radiation of wavelength \(242 \mathrm{~nm}\) is just sufficient to ionise the sodium atom. Calculate the ionisation energy of sodium in \(\mathrm{kJ} \mathrm{~mol}^{-1}\).

Answer:

\(
\begin{aligned}
& \mathrm{E}=h c / \lambda \\
& \lambda=242 \mathrm{~nm}=242 \times 10^{-9} \mathrm{~m}, c=3 \times 10^8 \mathrm{~m} \mathrm{~s}^{-1}, h=6.626 \times 10^{-34} \mathrm{Js}
\end{aligned}
\)
\(
E=\frac{\left(6.626 \times 10^{-34} \mathrm{Js}\right) \times\left(3 \times 10^8 \mathrm{~m} \mathrm{~s}^{-1}\right)}{\left(242 \times 10^{-9} \mathrm{~m}\right)}=0.0821 \times 10^{-17} \mathrm{~J}
\)
\(
\text { Ionisation energy per mol }(E)=\frac{\left(0 \cdot 0821 \times 10^{-17} \mathrm{~J}\right) \times\left(6.022 \times 10^{23} \mathrm{~mol}^{-1} \mathrm{~J}\right)}{1000}=\mathbf{4 9 4} \mathbf{~ k J ~ m o l}^{-1}
\)

Q11. A 25-watt bulb emits monochromatic yellow light of wavelength of \(0.57 \mu \mathrm{m}\). Calculate the rate of emission of quanta per second.

Answer: Energy of one photon \((\mathrm{E})=h v=h c / \lambda\)
\(
h=6.626 \times 10^{-34} \mathrm{~J} \mathrm{~s} ; c=3 \times 10^8 \mathrm{~m} \mathrm{~s}^{-1} ; \lambda=0.57 \times 10^{-6} \mathrm{~m}
\)
\(
\mathrm{E}=\frac{\left(6.626 \times 10^{-34} \mathrm{Js}\right) \times\left(3 \times 10^8 \mathrm{~m} \mathrm{~s}^{-1}\right)}{\left(0.57 \times 10^{-6} \mathrm{~m}\right)}=\mathbf{3 . 4 8} \times \mathbf{1 0}^{-19} \mathbf{J}
\)
\(
\text { Rate of emission of quanta per second }=\frac{\text { Power }}{\text { Energy }}
\)
\(
\text { Power }(P)=25 \text { watt }=25 \mathrm{~Js}^{-1} ; \mathrm{E}=3.48 \times 10^{-19} \mathrm{~J}
\)
\(
=\frac{(25 \text { watt })}{\left(3.48 \times 10^{-19} \mathrm{~J}\right)}=\frac{\left(25 \mathrm{Js}^{-1}\right)}{\left(3.48 \times 10^{-19} \mathrm{~J}\right)}=\mathbf{7 . 1 8} \times 10^{19} \mathrm{~s}^{-1} .
\)

Q12. Electrons are emitted with zero velocity from a metal surface when it is exposed to radiation of wavelength 6800 Å. Calculate threshold frequency \(\left(f_0\right)\) and work function \(\left(\mathrm{W}_0\right)\) of the metal.

Answer:

\(
\text { Threshold frequency }\left(f_0\right)=\frac{c}{\lambda}=\frac{\left(3 \times 10^8 \mathrm{~m} \mathrm{~s}^{-1}\right)}{\left(68 \times 10^{-8} \mathrm{~m}\right)}=\mathbf{4 . 4 1} \times 10^{14} \mathrm{~s}^{-1}
\)
\(
\text { Work function }\left(\mathrm{W}_0\right)=h f_0=\left(6.626 \times 10^{-34} \mathrm{Js}\right) \times\left(4.41 \times 10^{14} \mathrm{~s}^{-1}\right)=\mathbf{2 . 9 2} \times \mathbf{1 0}^{\mathbf{- 1 9}} \mathbf{J} \text {. }
\)

Q13. What is the wavelength of light emitted when the electron in a hydrogen atom undergoes the transition from an energy level with \(n=4\) to an energy level with \(n=2\)?

Answer:
According to the Balmer formula,
Wave number \((\tilde{\nu})=\mathrm{R}_{\mathrm{H}}\left[\frac{1}{n_1^2}-\frac{1}{n_2^2}\right] \mathrm{cm}^{-1} ; n_1=2, n_2=4, \mathrm{R}_{\mathrm{H}}=109678 \mathrm{~cm}^{-1}\)
\(
\begin{aligned}
& (\tilde{\nu})=109678\left[\frac{1}{2^2}-\frac{1}{4^2}\right] \mathrm{cm}^{-1}=\frac{109678 \times 3}{16} \mathrm{~cm}^{-1} . \\
& \lambda=\frac{1}{\tilde{\nu}}=\frac{16}{109678 \times 3} \mathrm{~cm}=\frac{16 \times 10^7}{109678 \times 3} \mathrm{~nm}=\mathbf{4 8 6} \mathrm{nm}
\end{aligned}
\)

Q14. How much energy is required to ionise a \(\mathrm{H}\) atom if the electron occuples \(n=5\) orbit? Compare your answer with the ionization enthalpy of \(\mathrm{H}\) atom ( energy required to remove the electron from \(n=1\) orbit).

Answer: Energy for a hydrogen electron present in a particular energy shell,
\(
\begin{aligned}
\mathrm{E}_n & =-\frac{13 \cdot 12}{n^2} \times 10^5 \mathrm{~J} \mathrm{~mol}^{-1}=-\frac{13 \cdot 12 \times 10^5}{n^2 \times 6 \cdot 022 \times 10^{23}} \mathrm{~J} \mathrm{atom}^{-1} \\
& =\frac{-2 \cdot 18 \times 10^{-18}}{n^2} \mathrm{~J} \mathrm{atom}^{-1}
\end{aligned}
\)
Step I. Ionisation energy for hydrogen electron present in orbit \(n=5\)
\(\mathrm{IE}_5=\mathrm{E}_{\infty}-\mathrm{E}_5=0-\left(\frac{-2.18 \times 10^{-18}}{25}\right) \mathrm{J} \text { atom }^{-1}=\mathbf{8 . 7 2} \times \mathbf{1 0}^{-\mathbf{2 0}} \mathrm{J} \text { atom }^{-1}\)
Step II. Ionisation energy for hydrogen electron present in orbit \(n=1\).
\(
\mathrm{IE}_1=\mathrm{E}_{\infty}-\mathrm{E}_1=0-\left(\frac{-2.18 \times 10^{-18}}{1}\right)=2.18 \times 10^{-18} \mathrm{~J} \mathrm{~atom}^{-1}
\)
On comparing :
\(
\frac{\mathrm{IE}_1}{\mathrm{IE}_5}=\frac{\left(2.18 \times 10^{-18} \mathrm{J} \text { atom }^{-1}\right)}{\left(8.72 \times 10^{-20} \mathrm{J} \text { atom }^{-1}\right)}=\mathbf{2 5}
\)
The energy required to remove an electron from first orbit in a hydrogen atom is 25 times the energy needed to remove an electron from the fifth orbit.

Q15. What is the maximum number of emission lines when the excited electron of a \(\mathrm{H}\) atom in \(n=6\) drops to the ground state?

Answer: The maximum no. of emission lines \(=\frac{n(n-1)}{2}=\frac{6(6-1)}{2}=3 \times 5=15\)
\(
\begin{aligned}
&\text { The actual transitions which are taking place are as follows : }\\
&\begin{array}{|c|c|c|c|c|}
\hline n=6 \text { to } n=1 & n=5 \text { to } n=1 & n=4 \text { to } n=1 & n=3 \text { to } n=1 & n=2 \text { to } n=1 \\
\hline 6 \rightarrow 5 & 5 \rightarrow 4 & 4 \rightarrow 3 & 3 \rightarrow 2 & 2 \rightarrow 1 \\
\hline 6 \rightarrow 4 & 5 \rightarrow 3 & 4 \rightarrow 2 & 3 \rightarrow 1 & \\
\hline 6 \rightarrow 3 & 5 \rightarrow 2 & 4 \rightarrow 1 & & \\
\hline 6 \rightarrow 2 & 5 \rightarrow 1 & & & \\
\hline \begin{array}{c}
6 \rightarrow 1 \\
(5 \text { lines })
\end{array} & \text { (4 lines) } & \text { (3 lines) } & \text { (2 lines) } & \text { (1 line) } \\
\hline
\end{array}
\end{aligned}
\)

Q16. (i) The energy associated with the first orbit in the hydrogen atom is \(-2.18 \times 10^{-18} \mathrm{J} \text { atom }{ }^{-1}\). What is the energy associated with the fifth orbit?
(ii) Calculate the radius of Bohr’s fifth orbit for the hydrogen atom.

Answer: (i) For an electron, the energies in two orbits may be compared as:
\(
\frac{\mathrm{E}_1}{\mathrm{E}_2}=\left(\frac{n_2}{n_1}\right)^2; \left[\because \quad \mathrm{E}_n \propto \frac{1}{n^2}\right]
\)
\(
\text { According to available data : } n_1=1, \mathrm{E}_1=-2.17 \times 10^{-18} \mathrm{J} \text { atom }{ }^{-1}, n_2=5
\)
\(
\therefore \frac{\left(-2 \cdot 17 \times 10^{-18} \mathrm{J} \text { atom }{ }^{-1}\right)}{\mathrm{E}_2}=\left(\frac{5}{1}\right)^2=25
\)
\(
E_5=\frac{\left(-2.17 \times 10^{-18} \mathrm{~J} \mathrm{~atom}^{-1}\right)}{25}=-\mathbf{8 . 7 7} \times \mathbf{1 0 ^ { – 2 0 }} \mathrm{J} \mathrm{~atom}^{-1} \text {. }
\)
(ii) For hydrogen atom; \(r_n=0.529 \times n^2 Å\); \(r_5=0.529 \times(5)^2=13.225 Å=1.3225 \mathrm{~nm}\).

Q17. Calculate the wavenumber for the longest wavelength transition in the Balmer series of atomic hydrogen.

Answer: According to Balmer formula, \((\tilde{\nu})=\frac{1}{\lambda}=\mathrm{R}_{\mathrm{H}}\left[\frac{1}{n_1^2}-\frac{1}{n_2^2}\right]\)
In order that the wavelength \((\lambda)\) may be the maximum, wave number \((\tilde{\nu})\) must be the least. This is possible in case \(n_2-n_1\) is minimum. Now, for the Balmer series, \(n_1=2\) and \(n_2\) must be 3. Substituting these values in the Balmer formula,
\(
(\tilde{\nu})=\left(1.097 \times 10^7 \mathrm{~m}^{-1}\right)\left(\frac{1}{2^2}-\frac{1}{3^2}\right)=1.097 \times 10^7 \mathrm{~m}^{-1}\left(\frac{5}{36}\right)=\mathbf{1 . 5 2 3} \times \mathbf{1 0 ^ { 6 }} \mathbf{m}^{-1}
\)

Q18. What is the energy in joules, required to shift the electron of the hydrogen atom from the first Bohr orbit to the fifth Bohr orbit and what is the wavelength of the light emitted when the electron returns to the ground state? The ground state electron energy is \(-2.18 \times 10^{-11} \mathrm{ergs}\).

Answer: Step I. Calculation of energy required
The energy of electron \(\left(\mathrm{E}_n\right)=\frac{-2 \cdot 18 \times 10^{-11}}{n^2}\) ergs \(=\frac{-2 \cdot 18 \times 10^{-18} \mathrm{~J}}{n^2}\)
\(\left(\because 1 \mathrm{~J}=10^7\right.\) ergs \()\)
\(
\text { The energy in Bohr’s first orbit }\left(E_1\right)=\frac{-2 \cdot 18 \times 10^{-18} \mathrm{~J}}{(1)^2}=\frac{-2 \cdot 18 \times 10^{-18} \mathrm{~J}}{1}
\)
\(
\text { The energy in Bohr’s fifth orbit }\left(E_5\right)=\frac{-2 \cdot 18 \times 10^{-18} \mathrm{~J}}{(5)^2}=\frac{-2 \cdot 18 \times 10^{-18} \mathrm{~J}}{25}
\)
\(
\text { Energy required }(\Delta \mathrm{E})=\mathrm{E}_5-\mathrm{E}_1=\left(\frac{-2 \cdot 18}{25} \times 10^{-18} \mathrm{~J}\right)-\left(-\frac{2 \cdot 18}{1} \times 10^{-18} \mathrm{~J}\right)
\)
\(
\begin{aligned}
& =2 \cdot 18 \times 10^{-18}\left(1-\frac{1}{25}\right) \mathbf{J} \\
& =2 \cdot 18 \times 10^{-18} \times 24 / 25=\mathbf{2} \cdot \mathbf{0 9} \times \mathbf{1 0}^{-18} \mathbf{J}
\end{aligned}
\)
Step II. Calculation of wavelength of light emitted
\(
\Delta \mathrm{E}=h f=\frac{h c}{\lambda}
\)
\(
\lambda=\frac{h c}{\Delta \mathrm{E}}=\frac{\left(6.626 \times 10^{-34} \mathrm{~J}-\mathrm{s}\right) \times\left(3 \times 10^8 \mathrm{~ms}^{-1}\right)}{\left(2.09 \times 10^{-18} \mathrm{~J}\right)}
\)
\(
=9.50 \times 10^{-8} \mathrm{~m}=950 Å \left(\because 1 Å=10^{-10} \mathrm{~m}\right)
\)

Q19. The electron energy in the hydrogen atom is given by \(E_n=\left(-2.18 \times 10^{-18}\right) / n^2 \mathrm{~J}\). Calculate the energy required to remove an electron completely from the \(n=2\) orbit. What is the longest wavelength of light in \(\mathrm{cm}\) that can be used to cause this transition?

Answer: Step I. Calculation of energy required
The energy required is the difference in the energy when the electron jumps from orbit with \(n=\infty\) to orbit with \(n=2\)
The energy required. \((\Delta E)=E_{\infty}-E_2\)
\(
=0-\left(-\frac{2.18 \times 10^{-18}}{4} \mathrm{~J}\right)=\mathbf{5 . 4 5} \times \mathbf{1 0 ^ { – 1 9 }} \mathbf{J}
\)
Step II. Calculation of the longest wavelength of light in \(\mathrm{cm}\) used to cause the transition
\(
\begin{aligned}
\Delta \mathrm{E} & =h f=h c / \lambda \\
\lambda & =\frac{h c}{\Delta \mathrm{E}}=\frac{\left(6.626 \times 10^{-34} \mathrm{Js}\right) \times\left(3 \times 10^8 \mathrm{~m} \mathrm{~s}^{-1}\right)}{\left(5.45 \times 10^{-19} \mathrm{~J}\right)} \\
& =3.644 \times 10^{-7} \mathrm{~m}=3.647 \times 10^{-7} \times 10^2=3.647 \times 10^{-5} \mathrm{~cm}
\end{aligned}
\)

Q20. Calculate the wavelength of an electron moving with a velocity of \(2.05 \times 10^7 \mathrm{~m} \mathrm{~s}^{-1}\).

Answer: According to de Broglie’s equation, \(\lambda=\frac{h}{m v}\)
\(
\begin{aligned}
\text { Mass of electron }(m) & =9.1 \times 10^{-31} \mathrm{~kg} \\
\text { Velocity of electron }(v) & =2.05 \times 10^7 \mathrm{~m} \mathrm{~s}^{-1} \\
\text { Planck’s constant }(h) & =6.626 \times 10^{-34} \mathrm{~kg} \mathrm{~m}^2 \mathrm{~s}^{-1}
\end{aligned}
\)
\(
\lambda=\frac{\left(6 \cdot 626 \times 10^{-34} \mathrm{~kg} \mathrm{~m}^2 \mathrm{~s}^{-1}\right)}{\left(9 \cdot 1 \times 10^{-31} \mathrm{~kg}\right) \times\left(2 \cdot 05 \times 10^7 \mathrm{~m} \mathrm{~s}^{-1}\right)}=\mathbf{3} \cdot \mathbf{5 5} \times \mathbf{1 0 ^ { – 1 1 }} \mathbf{m} .
\)

Q21. The mass of an electron is \(9.1 \times 10^{-31} \mathrm{~kg}\). If its K.E. is \(3.0 \times 10^{-25} \mathrm{~J}\), calculate its wavelength.

Answer: Step I. Calculation of velocity of the electron
\(
\text { Kinetic energy }=1 / 2 \mathrm{mv}^2=3.0 \times 10^{-25} \mathrm{~J}=3.0 \times 10^{-25} \mathrm{~kg} \mathrm{~m}^2 \mathrm{~s}^{-2}
\)
\(
v^2=\frac{2 \times \mathrm{K} . \mathrm{E} .}{m}=\frac{2 \times\left(3.0 \times 10^{-25} \mathrm{~kg} \mathrm{~m}^2 \mathrm{~s}^{-2}\right)}{\left(9.1 \times 10^{-31} \mathrm{~kg}\right)}=65.9 \times 10^4 \mathrm{~m}^2 \mathrm{~s}^{-2}
\)
\(
v=\left(65.9 \times 10^4 \mathrm{~m}^2 \mathrm{~s}^{-2}\right)^{1 / 2}=8.12 \times 10^2 \mathrm{~m} \mathrm{~s}^{-1}
\)
Step II. Calculation of wavelength of the electron According to de Broglie’s equation,
\(
\lambda=\frac{h}{m v}=\frac{\left(6.626 \times 10^{-34} \mathrm{~kg} \mathrm{~m}^2 \mathrm{~s}^{-1}\right)}{\left(9.1 \times 10^{-31} \mathrm{~kg}\right) \times\left(8.12 \times 10^2 \mathrm{~m} \mathrm{~s}^{-1}\right)}
\)
\(
=0.08967 \times 10^{-5} \mathrm{~m}=8967 \times 10^{-10} \mathrm{~m}=\mathbf{8 9 6 7} Å \left(\because 1 Å=10^{-10} \mathrm{~m}\right)
\)

Q22. Which of the following are isoelectronic species i.e., those having the same number of electrons?
\(
\mathrm{Na}^{+}, \mathrm{K}^{+}, \mathrm{Mg}^{2+}, \mathrm{Ca}^{2+}, \mathrm{S}^{2-}, \mathrm{Ar}
\)

Answer: \(\mathrm{Na}^{+}\)and \(\mathrm{Mg}^{2+}\) are iso-electronic species (have 10 electrons) \(\mathrm{K}^{+}, \mathrm{Ca}^{2+}, \mathrm{Ar}, \mathrm{S}^{2-}\) are iso-electronic species (have 18 electrons)

Q23. (i) Write the electronic configurations of the following ions: (a) \(\mathrm{H}^{-}\)(b) \(\mathrm{Na}^{+}\)(c) \(\mathrm{O}^{2-}\) (d) \(\mathrm{F}\)
(ii) What are the atomic numbers of elements whose outermost electrons are represented by (a) \(3 s^1\) (b) \(2 p^3\) and (c) \(3 p^5\)?
(iii) Which atoms are indicated by the following configurations?
(a) \([\mathrm{He}] 2 \mathrm{~s}^1\)
(b) \([\mathrm{Ne}] 3 s^2 3 p^3\)
(c) [Ar] \(4 s^2 3 d^1\).

Answer: (i) (a) \(1 s^2\)
(b) \(1 s^2 2 s^2 2 p^6\)
(c) \(1 s^2 2 s^2 2 p^6\)
(d) \(1 s^2 2 s^2 2 p^6\).
(ii) (a) \(\mathrm{Na}(Z=11)\) has outermost electronic configuration \(=3 \mathrm{~s}^1\)
(b) \(\mathrm{N}(Z=7)\) has outermost electronic configuration \(=2 p^3\)
(c) \(\mathrm{Fe}(\mathrm{Z}=26)\) has outermost electronic configuration \(=3 \mathrm{~d}^6\)
(iii) (a) \(\mathrm{Li}\)
(b) \(P\)
(c) \(\mathrm{Sc}\)

Q24. What is the lowest value of \(n\) that allows \(g\) orbitals to exist?

Answer: For g-orbital, Azimuthal quantum number, \(l=4\) as:
\(
\begin{aligned}
& l=0 \Rightarrow s, \\
& l=1 \Rightarrow p, \\
& l=2 \Rightarrow d, \\
& l=3 \Rightarrow f, \\
& l=4 \Rightarrow g,
\end{aligned}
\)
as value of \(l\) is from 0 to (n-1) where \(n=\) principal quantum number
\(
\begin{aligned}
& \therefore 4=n-1 \\
& n=5
\end{aligned}
\)

Q25. An electron is in one of the \(3 d\) orbitals. Give the possible values of \(n, l\) and \(m_l\) for this electron.

Answer: For the \(3 d\) orbital:
Principal quantum number \((n)=3\)
Azimuthal quantum number \((l)=2\)
Magnetic quantum number \(\left(m_l\right)=-2,-1,0,1,2\)

Q26. An atom of an element contains 29 electrons and 35 neutrons. Deduce (i) the number of protons and (ii) the electronic configuration of the element.

Answer: No. of protons in a neutral atom \(=\) No. of electrons \(=29\) Electronic configuration \(=1 s^2 2 s^2 2 p^6 3 s^2 3 p^6 3 d^{10} 4 s^1\).

Q27. Give the number of electrons in the species \(\mathrm{H}_2^{+}, \mathrm{H}_2\) and \(\mathrm{O}_2^{+}\)

Answer:
\(\mathrm{H}_2^{+}\):
Number of electrons present in hydrogen molecule \(\left(\mathrm{H}_2\right)=1+1=2\)
\(\therefore\) Number of electrons in \(\mathrm{H}_2^{+}=2-1=1\)
\(\mathrm{H}_2\) :
Number of electrons in \(\mathrm{H}_2=1+1=2\)
\(\mathrm{O}_2^{+}\):
Number of electrons present in oxygen molecule \(\left(\mathrm{O}_2\right)=8+8=16\)
\(\therefore\) Number of electrons in \(\mathrm{O}_2^{+}=16-1=15\)

Q28. (i) An atomic orbital has \(n=3\). What are the possible values of \(l\) and \(\mathrm{m}_l\)?
(ii) List the quantum numbers ( \(m\), and \(l\) ) of electrons for \(3 d\) orbital.
(iii) Which of the following orbitals are possible? \(1 p, 2 s, 2 p\) and \(3 f\)

Answer: (i) For \(n=3; l=0,1\) and 2.
For \({l}=0 ; {m}_l=0\)
For \(l=1 ; m_l=+1,0,-1\)
For \({l}=2 ; m_l=-2,-1,0,+1,+2\)
(ii) For an electron in 3rd orbital; \(n=3; l=2; m_l\) can have any of the values \(-2,-1,0\), \(+1,+2\)
\(
\text { (iii) } 1 p \text { and } 3 f \text { orbitals are not possible. }
\)

Explanation: Among the given orbitals only \(2 s\) and \(2 p\) are possible. \(1 p\) and \(3 f\) cannot exist.
For \(\mathrm{p}\)-orbital, \({l}=1\). For a given value of \({n}, l\) can have values from zero to \((n-1)\).
\(\therefore\) For \(l\) is equal to 1, the minimum value of \(n\) is 2.
Similarly,
For f-orbital, \(l=3\).
For \(l=3\), the minimum value of \({n}\) is 4.
Hence, \(1 p\) and \(3 f\) do not exist.

Q29. Using \(s, p, d\) notations, describe the orbital with the following quantum numbers.
(a) \(n=1, l=0\);
(b) \(n=3; l=1\)
(c) \(n=4 ; l=2\);
(d) \(n=4 ; l=3\).

Answer: (a) \(n=1, l=0\) (Given)
The orbital is \(1 s\).
(b) For \(n=3\) and \(l=1\)
The orbital is \(3 p\).
(c) For \(n=4\) and \(l=2\)
The orbital is \(4 d\).
(d) For \(n=4\) and \(l=3\)
The orbital is \(4 f\).

Q30. Explain, giving reasons, which of the following sets of quantum numbers are not possible.
(a) \(\quad n=0, \quad l=0, \quad m_l=0, \quad m_s=+1 / 2\)
(b) \(\quad n=1, \quad l=0, \quad m_l=0, \quad m_s=-1 / 2\)
(c) \(\quad n=1, \quad l=1, \quad m_l=0, \quad m_s=+1 / 2\)
(d) \(\quad n=2, \quad l=1, \quad m_l=0, \quad m_s=-1 / 2\)
(e) \(\quad n=3, \quad l=3, \quad m_l=-3, \quad m_s=+1 / 2\)
(f) \(\quad n=3, \quad l=1, \quad m_l=0, \quad m_s=+1 / 2\)

Answer: (a) The set of quantum numbers is not possible because the minimum value of \(n\) can be \(l\) and not zero.
(b) The set of quantum numbers is possible.
(c) The set of quantum numbers is not possible because, for \(n=1, l\) can not be equal to 1. It can have 0 value.
(d) The set of quantum numbers is not possible because for \(l=0. {m_l}\) cannot be +1. It must be zero.
(e) The set of quantum numbers is not possible because, for \(n=3,l \neq 3\).
(f) The set of quantum numbers is possible.

Q31. How many electrons in an atom may have the following quantum numbers?
(a) \(n=4, m_s=-1 / 2\)
(b) \(n=3, l=0\)

Answer: (a) For \(n=4\)
Total number of electrons \(=2 n^2=2 \times 16=32\)
Half out of these will have \({m_s}=-1 / 2\)
\(\therefore\) Total electrons with \({m_s}(-1 / 2)=16\)
(b) For \(n=3\)
\({l}=0 ; m_l=0, m_s= +1 / 2,-1 / 2\left(\right.\) two \(\left.\mathrm{e}^{-}\right)\)
\(n=3,l=0\) indicates that the electrons are present in the 3s orbital. Therefore, the number of electrons having \({n}=3\) and \(l=0\) is 2.

Q32. Show that the circumference of the Bohr orbit for the hydrogen atom is an integral multiple of the de Broglie wavelength associated with the electron revolving around the orbit.

Answer: According to Bohr’s theory,
\(
\begin{aligned}
& m v r=\frac{n h}{2 \pi} \\
& 2 \pi r=\frac{n h}{m v} \\
& m v=\frac{n h}{2 \pi r} \dots(i) \\
&
\end{aligned}
\)
According to de Broglie equation,
\(
\lambda=\frac{h}{m v}
\)
\(
m v=\frac{h}{\lambda} \dots(ii)
\)
Comparing (i) and (ii),
\(
\frac{n h}{2 \pi r}=\frac{h}{\lambda} \quad \text { or } \quad 2 \pi r=n \lambda
\)
Thus, the circumference \((2 \pi r)\) of the Bohr orbit for the hydrogen atom is an integral multiple of the de Broglie wavelength.

Q33. What transition in the hydrogen spectrum would have the same wavelength as the Balmer transition \(n=4\) to \(n=2\) of \(\mathrm{He}^{+}\)spectrum?

Answer: \(
\text { For an atom, } (\tilde{\nu})=\frac{1}{\lambda}=\mathrm{R}_{\mathrm{H}} \mathrm{Z}^2\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)
\)
For \(\mathrm{He}^{+}\)spectrum : \(\mathrm{Z}=4, n_2=4, n_1=2\).
\(
\therefore \quad (\tilde{\nu})=\frac{1}{\lambda}=\mathrm{R}_{\mathrm{H}} \times 4\left(\frac{1}{2^2}-\frac{1}{4^2}\right)=\frac{3 \mathrm{R}_{\mathrm{H}}}{4} \dots(i)
\)
\(
\text { For hydrogen spectrum : } \quad (\tilde{\nu})=\frac{3 \mathrm{R}_{\mathrm{H}}}{4} \text { and } \mathrm{Z}=1
\)
\(
\therefore \quad \quad (\tilde{\nu})=\frac{1}{\lambda}=\mathrm{R}_{\mathrm{H}} \times 1\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)
\)
or \(\quad \mathrm{R}_{\mathrm{H}}\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)=\frac{3 \mathrm{R}_{\mathrm{H}}}{4} \quad\) or \(\quad \frac{1}{n_1^2}-\frac{1}{n_2^2}=\frac{3}{4}\)
This corresponds to \(n_1=1, n_2=2\) and means that the transition has taken place in the Lyman series from \(n=2\) to \(n=1\).

Q34. Calculate the energy required for the process \(\mathrm{He}^{+}(\mathrm{g}) \rightarrow \mathrm{He}^{2+}(\mathrm{g})+\mathrm{e}^{-}\)
\(
\text { The ionization energy for the } \mathrm{H} \text { atom in the ground state is } 2.18 \times 10^{-18} \mathrm{~J} \mathrm{~atom}^{-1}
\)

Answer: The expression for the ionisation energy atom :
\(
\mathrm{E}_n=\frac{2 \cdot 18 \times 10^{-18} \times \mathrm{Z}^2}{n^2} \mathrm{J} \text { atom }{ }^{-1}
\)
For \(\mathrm{H}\) atom \((\mathrm{Z}=1), \mathrm{E}_{\mathrm{n}}=2.18 \times 10^{-18} \times(\mathrm{1})^2 \mathrm{J} \text { atom }{ }^{-1}\) (given)
For \(\mathrm{He}^{+}\)ion \((Z=2), E_n=2.18 \times 10^{-18} \times(2)^2=8.72 \times 10^{-18} \mathrm{J} \text { atom }{ }^{-1}\) (one electron species)

Q35. If the diameter of a carbon atom is \(0.15 \mathrm{~nm}\), calculate the number of carbon atoms that can be placed side by side in a straight line across the length of the scale of length \(20 \mathrm{~cm}\) long.

Answer: Length of scale \(=20 \mathrm{~cm}=20 \times 10^7 \mathrm{~nm}=2 \times 10^8 \mathrm{~nm}\)
Diameter of carbon atom \(=0.15 \mathrm{~nm}\)
\(\begin{aligned} \therefore \quad \text { Number of carbon atoms which can be placed side by side in the scale }= & \frac{\left(2 \times 10^8 \mathrm{~nm}\right)}{(0.15 \mathrm{~nm})} \\ & =1.33 \times 10^9\end{aligned}\)

Q36. \(2 \times 10^8\) atoms of carbon are arranged side by side. Calculate the radius of the carbon atom if the length of this arrangement is \(2.4 \mathrm{~cm}\).

Answer: The length of the arrangement \(=2.4 \mathrm{~cm}\)
Total number of carbon atoms present \(=2 \times 10^8\)
\(\begin{aligned} \text { Diameter of each carbon atom } & =\frac{(2.4 \mathrm{~cm})}{\left(2 \times 10^8\right)}=1.2 \times 10^{-8} \mathrm{~cm} \\ \text { Radius of each carbon atom } & =\frac{1}{2}\left(1.2 \times 10^{-8}\right)=6.0 \times 10^{-9} \mathrm{~cm}=0.06 \mathrm{~nm}\end{aligned}\)

Q37. The diameter of a zinc atom is 2.6 Å. Calculate (a) the radius of the zinc atom in pm and (b) the number of atoms present in a length of \(1.6 \mathrm{~cm}\) if the \(\mathrm{zinc}\) atoms are arranged side by side lengthwise.

Answer:

\(
\begin{aligned}
& \text { (a) Radius of zinc atom }=\frac{\text { Diameter }}{2} \\
& =\frac{2.6 Å}{2} \\
& =1.3 \times 10^{-10} \mathrm{~m} \\
& =130 \times 10^{-12} \mathrm{~m}=130 \mathrm{pm}
\end{aligned}
\)
(b) Length of the arrangement \(=1.6 \mathrm{~cm}\)
\(
=1.6 \times 10^{-2} \mathrm{~m}=1.6 \times 10^{10} \mathrm{pm}
\)
Diameter of zinc atom \(=2.6 \times 10^{-10} \mathrm{~m}=260 \mathrm{pm}\)
\(
\therefore \text { No. of zinc atoms placed side by side }=\frac{\left(1.6 \times 10^{10} \mathrm{pm}\right)}{(260 \mathrm{pm})}=\mathbf{6} \cdot \mathbf{1 5 } \times \mathbf{1 0 ^ { 7 }}
\)

Q38. A certain particle carries \(2.5 \times 10^{-16} \mathrm{C}\) of static electric charge. Calculate the number of electrons present in it.

Answer: Magnitude of charge \((q)=2.5 \times 10^{-16} \mathrm{C}\)
Charge on one electron \((e)=1.602 \times 10^{-19} \mathrm{C}\)
No. of electrons present \(=\frac{\left(2.5 \times 10^{-16} \mathrm{C}\right)}{\left(1.602 \times 10^{-19} \mathrm{C}\right)}=\mathbf{1 5 6 0}\)

Q39. In Milikan’s experiment, the static electric charge on the oll drops has been obtained by shining X-rays. If the static electric charge on the oil drop is \(-1.282 \times 10^{-18} \mathrm{C}\), calculate the number of electrons present on it.

Answer: Charge on oil droplet \(=-1.282 \times 10^{-18} \mathrm{C}\)
Charge on an electron \(=-1.602 \times 10^{-19} \mathrm{C}\)
Number of electrons \(=\frac{q}{e}=\frac{\left(-1.282 \times 10^{-18} \mathrm{C}\right)}{\left(-1.602 \times 10^{-19} \mathrm{C}\right)}=\mathbf{8}\)

Q40. In Rutherford’s experiment, generally, the thin foil of heavy atoms, like gold, platinum, etc. have been used to be bombarded by the \(\alpha\)-particles. If the thin foil of light atoms like aluminium etc. is used, what difference would be observed from the above results?

Answer: More number of \(\alpha\)-particles will pass as the nucleus of the lighter atoms is small, smaller number of \(\alpha\)-particles will be deflected as a number of positive charges is less than on the lighter nuclei.

Explanation: The difference that would be observed in the results of Rutherford’s experiment if a thin foil of light atoms like Aluminum is used:

• A foil of small atoms like Aluminum, having a very small number of positive charges will not be able to repulse a fast-moving alpha particle to return and at the maximum, it may deflect them at small angles.
• The number of \(\alpha\)-particles are deflected through small angles because of the large positive charge on the nucleus, hence If light atoms are used, their nuclei will be light, and moreover, they will have a small positive charge on the nucleus.
• Hence, in this observation, the return of alpha particles would not have been observed and the establishment of a nucleus would not have been made.

Q41. Symbols \({ }_{35}^{79} \mathrm{Br}\) and \({ }^{79} \mathrm{Br}\) can be written, whereas symbols \({ }_{79}^{35} \mathrm{Br}\) and \({ }^{35} \mathrm{Br}\) are not acceptable. Answer briefly.

Answer: For a given element the number of protons is the same for the isotopes, whereas the mass number can be different for the given atomic number.
The general convention of representing an element along with its atomic mass
(A) and atomic number \((Z)\) is \({ }_Z^A X\)
The atomic number of the element can be identified from its symbol because no two elements can have the atomic number. However, the mass numbers have to be mentioned in order to identify the elements. Thus, Symbols \({ }_{35}^{79} \mathrm{Br}\) and \({ }^{79} \mathrm{Br}\) are accepted because atomic number of \(\mathrm{Br}\) will remain 35 even if not mentioned. Symbol \({ }_{79}^{35} \mathrm{Br}\) is not accepted because atomic number of \(\mathrm{Br}\) cannot be 79 (more than the mass number = 35). Similarly, symbol \(35 \mathrm{Br}\) cannot be accepted because the mass number has to be mentioned. This is needed to differentiate the isotopes of an element.

Q42. An element with mass number 81 contains \(31.7 \%\) more neutrons as compared to protons. Assign the atomic symbol.

Answer: Let the number of protons in the element be \(\mathrm{x}\).
\(\therefore\) Number of neutrons in the element
\(
\begin{aligned}
& =x+31.7 \% \text { of } x \\
& =x+0.317 x \\
& =1.317 x
\end{aligned}
\)
According to the question,
The mass number of the element \(=81\)
\(\therefore\) (Number of protons + number of neutrons \()=81\)
\(
\begin{aligned}
& \Rightarrow x+1.317 x=81 \\
& 2.317 x=81 \\
& x=\frac{81}{2.317} \\
& =34.95 \\
& \therefore x=35
\end{aligned}
\)
Hence, the number of protons in the element i.e., \({x}\) is 35.
Since the atomic number of an atom is defined as the number of protons present in its nucleus, the atomic number of the given element is 35.
\(\therefore\) The atomic symbol of the element is \({ }_{35}^{81} \mathrm{Br}\)

Q43. An ion with mass number 37 possesses one unit of negative charge. If the ion contains \(11.1 \%\) more neutrons than the electrons, find the symbol of the ion.

Answer: Suppose the number of electrons in the ion \(=x\)
The number of neutrons \(=x+\frac{11.1}{100} x=1.111 x\)
Number of electrons in the neutral atom \(=x-1\)
\(\therefore\) Number of protons \(=x-1\)
Mass number \(=\) Number of neutrons + Number of protons
\(\therefore 37=1.111 x+x-1\) or \(2.111 x=38\) or \(x=18\)
\(\therefore\) Number of protons \(=\) Atomic no. \(=x-1=18-1=17\)
No. of electrons \(=18 ;\) No. of protons \(=17\)
Hence, the symbol of the ion will be \({ }_{17}^{37} \mathrm{Cl}^{-1}\).

Q44. An ion with mass number 56 contains 3 units of positive charge and \(30.4 \%\) more neutrons than electrons. Assign the symbol to this ion.

Answer: Let the no. of electrons in the ion \(=x\)
\(\therefore\) the no. of the protons \(=x+3\) (as the ion has three units positive charge)
and the no. of neutrons \(=x+\frac{x \times 30.4}{100}=\mathrm{x}+0.304 \mathrm{x}\)
Now, mass no. of ion \(=\) No. of protons + No. of neutrons
\(
\begin{aligned}
& =(x+3)+(x+0.304 x) \\
& \therefore 56=(x+3)+(x+0.304 x) \text { or } 2.304 x=56-3=53 \\
& x=\frac{53}{2.304}=23
\end{aligned}
\)
Atomic no. of the ion (or element) \(=23+3=26\)
The element with atomic number 26 is iron ( \(\mathrm{Fe}\) ) and the corresponding ion is \({ }_{26}^{56} \mathrm{Fe}^{3+}\)

Q45. Arrange the following types of radiations in increasing order of frequency: (a) radiation from a microwave oven (b) amber light from a traffic signal (c) radiation from FM radio (d) cosmic rays from outer space and (e) X-rays.

Answer: The increasing order of frequency is as follows:
\(
\text { Cosmic rays }>\mathrm{X} \text {-rays }>\text { amber colour }>\text { microwave }>\text { FM }
\)
The increasing order of wavelength is as follows:
\(
\text { Cosmic rays }<\mathrm{X} \text {-rays }<\text { amber colour }<\text { microwave }<\text { FM }
\)

Q46. Nitrogen laser produces a radiation at a wavelength of \(337.1 \mathrm{~nm}\). If the number of photons emitted is \(5.6 \times 10^{24}\), calculate the power of this laser.

Answer:

\(
\text { Power of the laser }(\mathrm{E})=\mathrm{N} h f=\mathrm{N} h \mathrm{c} / \lambda
\)
\(
=\frac{\left(5.6 \times 10^{24}\right) \times\left(6.626 \times 10^{-34} \mathrm{Js}\right) \times\left(3 \times 10^8 \mathrm{~m} \mathrm{~s}^{-1}\right)}{\left(337.1 \times 10^{-9} \mathrm{~m}\right)}=\mathbf{3 . 3} \times \mathbf{1 0}^6 \mathbf{J}
\)

Q47. Neon gas is generally used in the sign boards. If it emits strongly at \(616 \mathrm{~nm}\), calculate (a) the frequency of emission, (b) distance traveled by this radiation in \(30 \mathrm{~s}\) (c) energy of quantum and (d) number of quanta present if it produces \(2 \mathrm{~J}\) of energy.

Answer:

\(\text { (a) Frequency of emission (f) }=\frac{c}{\lambda}=\frac{\left(3.0 \times 10^8 \mathrm{~ms}^{-1}\right)}{\left(616 \times 10^{-9} \mathrm{~m}\right)}=\mathbf{4 . 8 7} \times \mathbf{1 0}^{{1 4}} \mathrm{s}^{-\mathbf{1}}\)
(b) Velocity of radiation \((c)=3.0 \times 10^8 \mathrm{~m} \mathrm{~s}^{-1}\)
Distance travelled in \(30 \mathrm{~s}=\left(3.0 \times 10^8 \mathrm{~m} \mathrm{~s}^{-1}\right) \times(30 \mathrm{~s})=\mathbf{9 . 0} \times \mathbf{1 0} \mathbf{~ m}\)
(c)
\(
\begin{aligned}
\text { Energy of quanta }(\mathrm{E}) & =h f=\frac{h c}{\lambda}=\frac{\left(6.626 \times 10^{-34} \mathrm{Js}\right) \times\left(3 \times 10^8 \mathrm{~m} \mathrm{~s}^{-1}\right)}{\left(616 \times 10^{-9} \mathrm{~m}\right)} \\
& =\mathbf{3 2 . 2 7} \times \mathbf{1 0}^{-\mathbf{2 0}} \mathbf{J}
\end{aligned}
\)
(d) Number of quanta present in \(2 \mathrm{~J}\) of energy
\(
=\frac{\text { Total energy }}{\text { Energy per quanta }}=\frac{(2 \mathrm{~J})}{\left(32.27 \times 10^{-20} \mathrm{~J}\right)}=\mathbf{6 . 2} \times 10^{18}
\)

Q48. In astronomical observations, signals observed from the distant stars are generally weak. If the photon detector receives a total of \(3.15 \times 10^{-18} \mathrm{~J}\) from the radiations of \(600 \mathrm{~nm}\), calculate the number of photons received by the detector.

Answer: From the expression of the energy of one photon (E),
\(
E=\frac{h c}{\lambda}
\)
Where, \(\lambda=\) wavelength of radiation
\({h}=\) Planck’s constant
\(c=\) velocity of radiation
Substituting the values in the given expression of \(E\) :
\(
E=\frac{\left(6.626 \times 10^{-34} \mathrm{Js}\right)\left(3 \times 10^8 \mathrm{~ms}^{-1}\right)}{\left(600 \times 10^{-9} \mathrm{~m}\right)}=3.313 \times 10^{-19} \mathrm{~J}
\)
Energy of one photon \(=3.313 \times 10^{-19} \mathrm{~J}\)
Number of photons received with \(3.15 \times 10^{-18} \mathrm{~J}\) energy
\(
\begin{aligned}
& \frac{3.15 \times 10^{-18} \mathrm{~J}}{3.313 \times 10^{-19} \mathrm{~J}} \\
& =9.5 \\
& \approx 10
\end{aligned}
\)

Q49. Lifetimes of the molecules in the excited states are often measured by using a pulsed radiation source of duration nearly in the nano-second range. If the radiation source has the duration of \(2 \mathrm{~ns}\) and the number of photons emitted during the pulse source is \(2.5 \times 10^{15}\), calculate the energy of the source.

Answer: Time duration \({t}=2 \mathrm{~ns}=2 \times 10^{-9} \mathrm{~s}\)
Frequency \((f)=\frac{1}{t}=\frac{1}{2 \times 10^{-9} \mathrm{~s}}=\frac{10^9}{2} \mathrm{~s}^{-1}\)
Energy of one photon, \(E=h f=\left(6.626 \times 10^{-34} \mathrm{Js}\right) \times\left(10^9 / 2 \mathrm{~s}^{-1}\right)=3.25 \times 10^{-25} \mathrm{~J}\) No. of photons \(=2.5 \times 10^5\)
\(\therefore\) Energy of source \(=3.3125 \times 10^{-25} \mathrm{~J} \times 2.5 \times 10^{15}=8.28 \times 10^{-10} \mathrm{~J}\)

Q50. The longest wavelength doublet absorption transition is observed at 589 and \(589.6 \mathrm{~nm}\). Calculate the frequency of each transition and energy difference between two excited states.

Answer:

\(
\begin{aligned}
& \lambda_1=589 \mathrm{~nm}=589 \times 10^{-9} \mathrm{~m} \\
& f_1=\frac{c}{\lambda_1}=\frac{3 \times 10^8 \mathrm{~m} \mathrm{~s}^{-1}}{589 \times 10^{-9} \mathrm{~m}}=\frac{3000}{589} \times 10^{14} \mathrm{~s}^{-1} \quad=\mathbf{5 . 0 9 3 4} \times 10^{14} \mathrm{~s}^{-1}
\end{aligned}
\)
\(
\begin{aligned}
& \lambda_2=589.6 \mathrm{~nm}=589.6 \times 10^{-9} \mathrm{~m} \\
& f_2=\frac{c}{\lambda_2}=\frac{3 \times 10^8 \mathrm{~m} \mathrm{~s}^{-1}}{589 \cdot 6 \times 10^{-9} \mathrm{~m}}=\frac{3000}{589 \cdot 6} \times 10^{14} \mathrm{~s}^{-1}=\mathbf{5 . 0 8 8 2} \times \mathbf{1 0}^{\mathbf{1 4}}
\end{aligned}
\)
\(
\Delta \mathrm{E}=\mathrm{E}_1-\mathrm{E}_2=\frac{h c}{\lambda_1}-\frac{h c}{\lambda_2}=h c\left[\frac{1}{\lambda_1}-\frac{1}{\lambda_2}\right]
\)
\(
=\left(6.626 \times 10^{-34} \mathrm{Js} \times 3 \times 10^8 \mathrm{~m} \mathrm{~s}^{-1}\right)\left[\frac{1}{589 \times 10^{-9} \mathrm{~m}}-\frac{1}{589.6 \times 10^{-9} \mathrm{~m}}\right]
\)
\(
=\frac{19 \cdot 878 \times 10^{-34} \times 10^8}{10^{-9}}\left[\frac{589 \cdot 6-589}{589 \cdot 6 \times 589}\right] \mathrm{J}
\)
\(
=\frac{19 \cdot 878 \times 10^{-17} \times 0.6}{589.6 \times 589} \mathbf{J}=\mathbf{3 . 4 5} \times \mathbf{1 0 ^ { – 2 2 }} \mathbf{J}
\)

Q51. The work function for caesium atom is \(1.9 \mathrm{eV}\). Calculate (a) the threshold wavelength and (b) the threshold frequency of the radiation. If the caesium element is irradiated with a wavelength \(500 \mathrm{~nm}\), calculate the kinetic energy and the velocity of the ejected photoelectron.

Answer:
(a)
\(
\mathrm{E}_0=1.9 \mathrm{eV}=1.9 \times 1.602 \times 10^{-19} \mathrm{~J}
\)
\(
\text { Threshold frequency }\left(f_0\right)=\frac{E_0}{h}=\frac{1.9 \times 1.602 \times 10^{-19} \mathrm{~J}}{6.626 \times 10^{-34} \mathrm{JS}}=0.459 \times 10^{15} \mathrm{~s}^{-1}=\mathbf{4 . 5 9} \times 10^{14} \mathrm{~s}^{-1}
\)
\(
\text { Threshold wavelength }\left(\lambda_0\right)=\frac{c}{v_0}=\frac{3 \times 10^8 \mathrm{~m} \mathrm{~s}^{-1}}{4.59 \times 10^{14} \mathrm{~s}^{-1}}=0.6536 \times 10^{-6} \mathrm{~m} \approx \mathbf{6 5 3} \mathbf{n m}
\)
\(
\text { Hence, the threshold wavelength } \lambda_0 \text { is } 653 \mathrm{~nm} \text {. }
\)
(b)

\(
\text { Threshold frequency }\left(v_0\right)=\frac{\mathrm{E}_0}{h}=\frac{1.9 \times 1.602 \times 10^{-19} \mathrm{~J}}{6.626 \times 10^{-34} \mathrm{Js}}=0.459 \times 10^{15} \mathrm{~s}^{-1}=\mathbf{4} \cdot \mathbf{5 9} \times 10^{14} \mathrm{~s}^{-1}
\)

The velocity of the Photoelectron is
\(
\begin{aligned}
\text { Velocity }(v) & =\sqrt{\frac{2 \times 9 \cdot 36 \times 10^{-20}}{m} \mathrm{~J}}=\sqrt{\frac{2 \times 9 \cdot 36 \times 10^{-20} \mathrm{~kg} \mathrm{~m}^2 \mathrm{~s}^{-2}}{9 \cdot 1 \times 10^{-31} \mathrm{~kg}}} \\
& =\sqrt{2.057 \times 10^{11} \mathrm{~m}^2 \mathrm{~s}^{-2}}=\sqrt{20 \cdot 57 \times 10^{10} \mathrm{~m}^2 \mathrm{~s}^{-2}}=\mathbf{4 . 5 3 5 6} \times \mathbf{1 0} \mathbf{m ~ s}^{-\mathbf{1}}
\end{aligned}
\)
(c)
\(
\mathrm{E}=\mathrm{E}_0+\frac{1}{2} m v^2
\)
\(
\text { Kinetic energy }\left(\frac{1}{2} m v^2\right)=\mathrm{E}-\mathrm{E}_0=h c\left[\frac{1}{\lambda}-\frac{1}{\lambda_0}\right]
\)
\(
\begin{aligned}
& =\frac{\left(6.626 \times 10^{-34} \mathrm{Js}\right) \times\left(3 \times 10^8 \mathrm{~m} \mathrm{~s}^{-1}\right)}{10^{-9} \mathrm{~m}} \times\left[\frac{1}{500}-\frac{1}{654}\right] \\
& =\frac{6.626 \times 3 \times 154}{500 \times 654} \times 10^{-34+8+9}=\mathbf{9 . 3 6} \times 10^{-20} \mathbf{J}
\end{aligned}
\)

Q52. The following results are observed when sodium metal is irradiated with different wavelengths. Calculate (a) threshold wavelength and, (b) Planck’s constant.
\(\begin{array}{llll}\lambda(\mathrm{nm}) & 500 & 450 & 400 \\ \mathrm{v} \times 10^{-5}\left(\mathrm{~cm} \mathrm{~s}^{-1}\right) & 2.55 & 4.35 & 5.35\end{array}\)

Answer: (a) Let threshold wavelength \(=\lambda_0 \mathrm{~nm}=\lambda_0 \times 10^{-9} \mathrm{~m}\) According to photoelectric effect :
\(
\begin{aligned}
h\left(f-f_0\right) & =\frac{1}{2} m v^2 \\
h c\left(\frac{1}{\lambda}-\frac{1}{\lambda_0}\right) & =\frac{1}{2} m v^2
\end{aligned}
\)
Substituting the results of three experiments in the above equation :
\(
\begin{aligned}
& \frac{h c}{10^{-9}}\left(\frac{1}{500}-\frac{1}{\lambda_0}\right)=\frac{1}{2} \mathrm{~m}\left(2.55 \times 10^5\right)^2 \dots(i) \\
& \frac{h c}{10^{-9}}\left(\frac{1}{450}-\frac{1}{\lambda_0}\right)=\frac{1}{2} \mathrm{~m}\left(4.35 \times 10^5\right)^2 \dots(ii) \\
& \frac{h c}{10^{-9}}\left(\frac{1}{400}-\frac{1}{\lambda_0}\right)=\frac{1}{2} \mathrm{~m}\left(5.35 \times 10^5\right)^2 \dots(iii)
\end{aligned}
\)
Divide eqn. (ii) by eqn. (i),
\(
\frac{\left(\lambda_0-450\right)}{450 \times \lambda_0} \times \frac{500 \times \lambda_0}{\left(\lambda_0-500\right)}=\frac{\left(4.35 \times 10^5\right)^2}{\left(2.99 \times 10^5\right)^2}
\)
\(
\frac{\left(\lambda_0-450\right)}{\left(\lambda_0-500\right)}=\frac{(4.35)^2}{(2.99)^2} \times \frac{450}{500}=2.619
\)
\(
\left(\lambda_0-450\right)=2 \cdot 619\left(\lambda_0-500\right)=2 \cdot 619 \lambda_0-1309 \cdot 5
\)
\(
1.619 \lambda_0=859.5 \therefore \lambda_0=\frac{859 \cdot 5}{1.619}=\mathbf{5 3 1} \mathrm{nm}
\)

(b)

\(
\begin{aligned}
&\text { The value of the threshold wavelength is substituted in equation (iii). }\\
&\begin{aligned}
& h \times 3 \times 10^8\left(\frac{1}{400 \times 10^{-9}}-\frac{1}{526 \times 10^{-9}}\right)=\frac{1}{2} \times 9.11 \times 10^{-31} \times\left(5.2 \times 10^5\right)^2 \\
& h=6.84 \times 10^{-34} Js
\end{aligned}
\end{aligned}
\)

Q53. The ejection of the photoelectron from the silver metal in the photoelectric effect experiment can be stopped by applying the voltage of \(0.35 \mathrm{~V}\) when the radiation \(256.7 \mathrm{~nm}\) is used. Calculate the work function for silver metal.

Answer:

\(
\begin{aligned}
\lambda & =256.7 \mathrm{~nm}=256.7 \times 10^{-9} \mathrm{~m} ; \text { K.E. }=0.35 \mathrm{~eV} \\
\mathrm{E} & =\frac{h c}{\lambda}=\frac{\left(6.626 \times 10^{-34} \mathrm{Js}\right)\left(3 \times 10^8 \mathrm{~m} \mathrm{~s}^{-1}\right)}{\left(256.7 \times 10^{-19} \mathrm{~m}\right)} \\
& =\frac{6.626 \times 3}{256.7} \times 10^{-17} \mathrm{~J}=\frac{6.626 \times 3 \times 10^{-17}}{256.7 \times 1.602 \times 10^{-19}} \mathrm{~eV} \\
& =\frac{662.6 \times 3}{256.7 \times 1.602} \mathrm{eV}=4.83 \mathrm{~eV}
\end{aligned}
\)
\(
\begin{aligned}
\mathrm{E} & =\mathrm{E}_0+\mathrm{K} . \mathrm{E} . \\
4.83 \mathrm{~eV} & =\mathrm{E}_0+0.35 \mathrm{~eV} \\
\mathrm{E}_0 & =4.83-0.35=\mathbf{4 . 4 8} \mathrm{~eV}
\end{aligned}
\)

Q54. If the photon of the wavelength \(150 \mathrm{pm}\) strikes an atom and one of its inner bound electrons is ejected out with a velocity of \(1.5 \times 10^7 \mathrm{~m} \mathrm{~s}^{-1}\), calculate the energy with which it is bound to the nucleus.

Answer:

\(
\lambda=150 \mathrm{pm}=150 \times 10^{-12} \mathrm{~m}=1.5 \times 10^{-10} \mathrm{~m} ; v=1.5 \times 10^7 \mathrm{~m} \mathrm{~s}^{-1}
\)
\(
\begin{aligned}
\text { K.E. } & =\frac{1}{2} m v^2=\frac{1}{2} \times 9.1 \times 10^{-31} \mathrm{~kg} \times\left(1.5 \times 10^7 \mathrm{~m} \mathrm{~s}^{-1}\right)^2 \\
& =\frac{9.1 \times 1.5 \times 1.5}{2} \times 10^{-31+14} \mathrm{~J} \\
& =10.2375 \times 10^{-17} \mathrm{~J}=1.02375 \times 10^{-16} \mathrm{~J}
\end{aligned}
\)
\(
\begin{aligned}
\mathrm{E} & =\frac{h c}{\lambda}=\frac{\left(6.626 \times 10^{-34} \mathrm{Js}\right)\left(3 \times 10^8 \mathrm{~m} \mathrm{~s}^{-1}\right)}{\left(1.5 \times 10^{-10} \mathrm{~m}\right)} \\
& =\frac{6.626 \times 3}{1.5} \times 10^{-34+8+10} \\
& =\frac{6.626 \times 3}{1.5} \times 10^{-16} \mathrm{~J}=13.252 \times 10^{-16} \mathrm{~J}
\end{aligned}
\)
\(
\begin{aligned}
E & =E_0+\text { K.E. } \\
E_0 & =E-K . E .=(13.252-1.024) \times 10^{-16} J
\end{aligned}
\)
\(
=12.362 \times 10^{-16} \mathrm{~J}=\frac{12.228 \times 10^{-16}}{1.602 \times 10^{-19}}=\mathbf{7 . 6 3} \times 10^3 \mathrm{eV}
\)

Q55. Emission transitions in the Paschen series end at orbit \(n=3\) and start from orbit \(n\) and can be represented as \(v=3.29 \times 10^{15}(\mathrm{~Hz})\left[1 / 3^2-1 / \mathrm{n}^2\right]\) Calculate the value of \(\mathrm{n}\) if the transition is observed at \(1285 \mathrm{~nm}\). Find the region of the spectrum.

Answer: \(
\begin{aligned}
& f=\left(3.29 \times 10^{15} \mathrm{~Hz}\right)\left(\frac{1}{3^2}-\frac{1}{n^2}\right) \\
& \lambda=1285 \mathrm{~nm}=1285 \times 10^{-19} \mathrm{~m}=1.285 \times 10^{-16} \mathrm{~m} \\
& f=\frac{c}{\lambda}=\frac{\left(3 \times 10^8 \mathrm{~m} \mathrm{~s}^{-1}\right)}{\left(1.285 \times 10^{-6} \mathrm{~m}\right)}=2.3346 \times 10^{14} \mathrm{~s}^{-1}
\end{aligned}
\)
\(
2.3346 \times 10^{14}=3.29 \times 10^{15}\left[\frac{1}{3^2}-\frac{1}{n^2}\right]
\)
\(
\frac{2 \cdot 3346}{32 \cdot 9}=\frac{1}{3^2}-\frac{1}{n^2} \text { or } 0 \cdot 071=\frac{1}{9}-\frac{1}{n^2}
\)
\(
\begin{aligned}
\frac{1}{n^2} & =\frac{1}{9}-0.071=0.111-0.071=0.04 \\
n^2 & =\frac{1}{0.04}=25 \text { or } \boldsymbol{n}=\mathbf{5}
\end{aligned}
\)
Paschen series lies in the infrared region of the spectrum.

Q56. Calculate the wavelength for the emission transition if it starts from the orbit having radius \(1.3225 \mathrm{~nm}\) and ends at \(211.6 \mathrm{pm}\). Name the series to which this transition belongs and the region of the spectrum.

Answer:

\(
\text { Radius of the orbit of } \mathrm{H} \text { like species }=\frac{0.529}{\mathrm{Z}} n^2 Å=\frac{52.9}{\mathrm{Z}} n^2 \mathrm{pm}
\)
\(
\begin{aligned}
& r_1=1.3225 \mathrm{~nm}=1322.5 \mathrm{pm}=\frac{52.9 n_1^2}{\mathrm{Z}} \dots(i) \\
& r_2=211.6 \mathrm{pm}=\frac{52.9 n_2^2}{\mathrm{Z}} \dots(ii)
\end{aligned}
\)
\(
\begin{aligned}
\frac{r_1}{r_2} & =\frac{1322 \cdot 5}{211 \cdot 6}=\frac{n_1^2}{n_2^2} \\
\frac{n_1^2}{n_2^2} & =6.25 \text { or } \frac{n_1}{n_2}=(6.25)^{1 / 2}=2.5
\end{aligned}
\)
Thus, if \(n_1=2, n_2=5\). This transition corresponds to the transition from 5 th orbit to the 2nd orbit. This means that the transition belongs to the Balmer series.
Now,
\(
\begin{aligned}
\text { Wave number }(\tilde{\nu}) & =\left(1.097 \times 10^7 \mathrm{~m}^{-1}\right) \times\left(\frac{1}{2^2}-\frac{1}{5^2}\right) ;=1.097 \times 10^7 \times \frac{21}{100} \mathrm{~m}^{-1} \\
& =23.037 \times 10^5 \mathrm{~m}^{-1} \\
\text { Wavelength }(\lambda) & =\frac{1}{\bar{v}}=\frac{1}{\left(23.037 \times 10^5 \mathrm{~m}^{-1}\right)}=434 \times 10^{-9} \mathrm{~m} \text { or } 434 \mathrm{~nm}
\end{aligned}
\)
It lies in the visible region of light.

Q57. The dual behaviour of matter proposed by de Broglie led to the discovery of an electron microscope often used for the highly magnified images of biological molecules and other types of material. If the velocity of the electron in this microscope is \(1.6 \times 10^6\) \(\mathrm{ms}^{-1}\), calculate de Broglie wavelength associated with this electron.

Answer:

\(
\lambda=\frac{h}{m \cdot v}=\frac{\left(6.626 \times 10^{34} \mathrm{Js}\right)}{\left(9.1 \times 10^{-31} \mathrm{~kg}\right) \times\left(1.6 \times 10^6 \mathrm{~m} \mathrm{~s}^{-1}\right)}
\)
\(
=0.455 \times 10^{-34+25} \mathrm{~m}=0.455 \mathrm{~nm}=\mathbf{4 5 5} \mathrm{pm}
\)

Q58. Similar to electron diffraction, neutron diffraction microscope is also used for the determination of the structure of molecules. If the wavelength used here is \(800 \mathrm{pm}\), calculate the characteristic velocity associated with the neutron.

Answer: 

\(
\begin{aligned}
\lambda & =800 \mathrm{pm}=800 \times 10^{-12} \mathrm{~m}=8 \times 10^{-10} \mathrm{~m} \\
m & =1.675 \times 10^{-27} \mathrm{~kg}
\end{aligned}
\)
\(
v=\frac{h}{m \lambda}=\frac{\left(6.626 \times 10^{-34} \mathrm{~kg} \mathrm{~m}^2 \mathrm{~s}^{-1}\right)}{\left(1.675 \times 10^{-27} \mathrm{~kg}\right) \times\left(8 \times 10^{-10} \mathrm{~m}\right)}
\)
\(
=\frac{6.626}{1.675 \times 8} \times 10^{-34+37} \mathrm{~ms}^{-1}=\frac{6.626 \times 10^3}{1.675 \times 8} \mathrm{~ms}^{-1}
\)
\(
=0.494 \times 10^3 \mathrm{~ms}^{-1}=494 \mathrm{~m} \mathrm{~s}^{-1} .
\)

Q59. If the velocity of the electron in Bohr’s first orbit is \(2.19 \times 10^6 \mathrm{~ms}^{-1}\), calculate the de Broglie wavelength associated with it.

Answer: 

\(
v=2.19 \times 10^6 \mathrm{~m} \mathrm{~s}^{-1}
\)
\(
\lambda=\frac{h}{m v}=\frac{\left(6.626 \times 10^{-34} \mathrm{~kg} \mathrm{~m}^2 \mathrm{~s}^{-1}\right)}{\left(9.1 \times 10^{-31} \mathrm{~kg}\right) \times\left(2.19 \times 10^6 \mathrm{~ms}^{-1}\right)}
\)
\(
=\frac{6 \cdot 626}{9 \cdot 1 \times 2 \cdot 19} \times 10^{-34+25} \mathrm{~m}=0.33243 \times 10^{-9} \mathrm{~m}=332.43 \mathrm{pm} .
\)

Q60. The velocity associated with a proton moving in a potential difference of \(1000 \mathrm{~V}\) is \(4.37 \times 10^5 \mathrm{~ms}^{-1}\). If the hockey ball of mass \(0.1 \mathrm{~kg}\) is moving with this velocity, calculate the wavelength associated with this velocity.

Answer: 

\(
v=4.37 \times 10^9 \mathrm{~m} \mathrm{~s}^{-1} ; m=0.1 \mathrm{~kg}
\)
\(
\lambda=\frac{h}{m v}=\frac{\left(6.626 \times 10^{-34} \mathrm{~kg} \mathrm{~m}^2 \mathrm{~s}^{-1}\right)}{(0 \cdot 1 \mathrm{~kg}) \times\left(4.37 \times 10^5 \mathrm{~m} \mathrm{~s}^{-1}\right)}=\frac{6.626}{0.437} \times 10^{-34-5} \mathrm{~m}
\)
\(
=15.16 \times 10^{-39} \mathrm{~m}=\mathbf{1} \cdot \mathbf{5 1 6} \times \mathbf{1 0}^{-\mathbf{3 8}} \mathbf{m}
\)

Q61. If the position of the electron is measured within an accuracy of \(\pm 0.002 \mathrm{~nm}\), calculate the uncertainty in the momentum of the electron. Suppose the momentum of the electron is \(h / 4 \pi_{\mathrm{m}} \times 0.05 \mathrm{~nm}\), is there any problem in defining this value?

Answer:

\(
\begin{aligned}
\Delta x & =0.002 \mathrm{~nm}=0.002 \times 10^{-9} \mathrm{~m}=2.0 \times 10^{-12} \mathrm{~m} \\
\Delta x \cdot \Delta p & =\frac{h}{4 \pi} \text { or } \Delta p=\frac{h}{4 \pi \Delta x}
\end{aligned}
\)
\(
\Delta p=\frac{\left(6.626 \times 10^{-34} \mathrm{~kg} \mathrm{~m}^2 \mathrm{~s}^{-1}\right)}{4 \times 3.142 \times\left(2 \times 10^{-12} \mathrm{~m}\right)}=2.638 \times 10^{-23} \mathrm{~kg} \mathrm{~m} \mathrm{~s}^{-1}
\)
\(
\text { Actual momentum }(p)=\frac{h}{4 \pi \times 0 \cdot 05 \mathrm{~nm}}=\frac{\left(6 \cdot 626 \times 10^{-34} \mathrm{~kg} \mathrm{~m}^2 \mathrm{~s}^{-1}\right)}{4 \times 3 \cdot 142 \times\left(5 \times 10^{-11} \mathrm{~m}\right)}=\mathbf{1 . 0 5 5} \times \mathbf{1 0}^{-24} \mathbf{k g ~ m ~ s}^{\mathbf{- 1}}
\)
Since actual momentum is smaller than the uncertainty in measuring momentum, therefore, the momentum of an electron can not be defined.

Q62. The quantum numbers of six electrons are given below. Arrange them in order of increasing energies. If any of these combination(s) has/have the same energy lists:
1. \(n=4, l=2, m_{\mathrm{l}}=-2, m_{\mathrm{s}}=-1 / 2\)
2. \(n=3, l=2, m_l=1, m_{\mathrm{s}}=+1 / 2\)
3. \(n=4, l=1, m_l=0, m_{\mathrm{s}}=+1 / 2\)
4. \(n=3, l=2, m_l=-2, m_{\mathrm{s}}=-1 / 2\)
5. \(n=3, l=1, m_l=-1, m_{\mathrm{s}}=+1 / 2\)
6. \(n=4, l=1, m_{\mathrm{l}}=0, m_{\mathrm{s}}=+1 / 2\)

Answer: 

(i) \(4 d\)
(ii) \(3 d\)
(iii) \(4 p\)
(iv) \(3 d\)
(v) \(3 p\)
(vi) \(4 p\).
For \({n}=4\) and \(l=2\), the orbital occupied is \(4 d\).
For \(n=3\) and \(l=2\), the orbital occupied is \(3 d\).
For \(n=4\) and \(l=1\), the orbital occupied is \(4 p\).
Hence, the six electrons i.e., 1, 2, 3, 4, 5, and 6 are present in the \(4 d, 3 d, 4 p, 3 d\), \(3 p\), and \(4 p\) orbitals respectively.
Therefore, the increasing order of energies is
\(
\text { (v) }<\text { (ii) }=(\text { iv })<(\text { vi })=(\text { iii })<\text { (i) }
\)

Q63. The bromine atom possesses 35 electrons. It contains 6 electrons in \(2 p\) orbital, 6 electrons in \(3 p\) orbital and 5 electron in \(4 p\) orbital. Which of these electrons experiences the lowest effective nuclear charge?

Answer: The nuclear charge experienced by an electron (present in a multi-electron atom) is dependent upon the distance between the nucleus and the orbital, in which the electron is present. As the distance increases, the effective nuclear charge will decrease. Among \({p}\)-orbitals, \(4 {p}\) orbitals are farthest from the nucleus of the bromine atom with \((+35)\) charge. Hence, the electrons in the \(4 p\) orbital will experience the lowest effective nuclear charge. These electrons are shielded by electrons present in the \(2 p\) and \(3 p\) orbitals along with the s-orbitals. Therefore, they will experience the lowest nuclear charge.

Q64. Among the following pairs of orbitals which orbital will experience the larger effective nuclear charge? (i) \(2 s\) and \(3 s\), (ii) \(4 d\) and \(4 f\), (iii) \(3 d\) and \(3 p\).

Answer: Nuclear charge is defined as the net positive charge experienced by an electron in the orbital of a multi-electron atom. The closer the orbital, the greater is the nuclear charge experienced by the electron (s) in it. Please note that the greater the penetration of the electron present in a particular orbital towards the nucleus, more will be the magnitude of the effective nuclear charge. Based on this,
(i) The electron(s) present in the \(2 s\) orbital will experience greater nuclear charge (being closer to the nucleus) than the electron(s) in the \(3 s\) orbital
(ii) \(4 d\) will experience greater nuclear charge than \(4 f\) since \(4 d\) is closer to the nucleus.
(iii) \(3 p\) will experience greater nuclear charge since it is closer to the nucleus than \(3 d\).

Q65. The unpaired electrons in \(\mathrm{Al}\) and \(\mathrm{S} 1\) are present in \(3 p\) orbital. Which electrons will experience more effective nuclear charge from the nucleus?

Answer: Configuration of the two elements are:
\(\mathrm{A} 1(\mathrm{Z}=13):[\mathrm{Ne}]^{10} 3 \mathrm{~s}^2 3 \mathrm{p}^1 ; \mathrm{Si}(\mathrm{Z}=14):[\mathrm{Ne}] ^{10} 3 \mathrm{~s}^2 3 \mathrm{p}^2\)
The unpaired electrons in silicon (Si) will experience more effective nuclear charge because the atomic number of the element \(\mathrm{Si}\) is more than that of \(\mathrm{Al}\).

Explanation: Nuclear charge is defined as the net positive charge experienced by an electron in a multielectron atom. The higher the atomic number, the higher is the nuclear charge. Silicon has 14 protons while aluminium has 13 protons. Hence, silicon has a larger nuclear charge of \((+14)\) than aluminum, which has a nuclear charge of (+13). Thus, the electrons in the \(3 p\) orbital of silicon will experience a more effective nuclear charge than aluminum.

Q66. Indicate the number of unpaired electrons in (a) P, (b) Si, (c) \(\mathrm{Cr}\), (d) \(\mathrm{Fe}\) and (e) \(\mathrm{Kr}\).

Answer: (a) Phosphorus (P): \(1 s^2 2 s^2 2 p^6 3 s^2 3 p^3\)
Thus we can see that it has 5 valence electrons in its outermost shell.
Calculating the unpaired electrons:
In Phosphorus, the p-subshell has 3 electrons in it.
Now we know that these electrons will fill the orbitals according to Hund’s rule of maximum multiplicity
That is the orbitals will first be filled singly and after all the orbitals are filled singly, then the pairing of the electrons will start.
For Phosphorus, if we draw the 3 s and \(3 p\) orbital:
Here we can see that in the \(3 p\) orbital, there are 3 unpaired electrons present. Therefore, Phosphorus has 3 unpaired electrons present in it.
No. of the unpaired electron \(=3\)
(b) Silicon (Si): \(1 s^2 2 s^2 2 p^6 3 s^2 3 p^2\)
No. of unpaired electron \(=2\) (since \(\mathrm{p}\) orbital can have maximum 6 electron )
(c) Chromium (Cr): \(1 s^2 2 s^2 2 p^6 3 s^2 3 p^6 4 s^1 3 d^5\)
No. of unpaired electrons \(=6\) (since 1 electron is to be added to 4 s & 5 electrons to be added to \(3 d\) orbital.
(d) Iron (Fe): \(1 s^2 2 s^2 2 p^6 3 s^2 3 p^6 4 s^2 3 d^6\)
No. of unpaired electrons \(=4\)
(e) Krypton (Kr): \(1 s^2 2 s^2 2 p^6 3 s^2 3 p^6 4 s^2 3 d^10 4 p^6\)
Since all orbitals are fully occupied, there are no unpaired (zero) electrons in krypton.

Q67. (a) How many subshells are associated with \(n=4\)? (b) How many electrons will be present in the subshells having \(\mathrm{m}_{\mathrm{s}}\) value of \(-1 / 2\) for \(n=4\)?

Answer: (a) For \(n=4\); No. of sub-shells \(=(l=0, l=1, l=2, l=3)=\),4 .
(b) Total number of orbitals that can be present \(=n^2=4^2=16\).
Each orbital can have an electron with \(m_s=-1 / 2 \therefore\) Total no. of electrons with \(m_s=-1 / 2\) is 16 .

Explanation: (a) Number of sub-shells associated with \(n=4\) :
– A given value of \(n, l\) can have values from zero to \((\mathrm{n}-1)\). i. e., \(0,1,2,3\)
– The four sub-shells are associated with \(\mathrm{n}=4\), are \(\mathrm{s}, \mathrm{p}, \mathrm{d}\), and \(\mathrm{f}\).
– Where,
\(
\begin{gathered}
l=0: 4 \mathrm{~s}-\text { subshell } \\
l=1: 4 \mathrm{p}-\text { subshell } \\
l=2: 4 \mathrm{~d}-\text { subshell } \\
l=3: 4 \mathrm{f}-\text { subshell }
\end{gathered}
\)

(b) The number of electrons in the sub-shells having an \(m_s\) value of \(-1 / 2\) for \(n\) \(=4\) :
– The number of orbitals will be \(=\mathrm{n}^2\)
– The number of orbitals \(=16\).
– Two electrons will go to each orbital with spins \(-1 / 2\) and \(+1 / 2\)
\(\therefore\) Hence the number of electrons having spin \(-1 / 2\) will be 16.

Q68. The arrangement of orbitals on the basis of energy is based upon their \((n+l)\) value. The lower the value of \((n+l)\), the lower is the energy. For orbitals having the same values of \((n+l)\), the orbital with the lower value of \(n\) will have lower energy.
I. Based upon the above information, arrange the following orbitals in the increasing order of energy.
(a) \(1 s, 2 s, 3 s, 2 p\)
(b) \(4 s, 3 s, 3 p, 4 d\)
(c) \(5 p, 4 d, 5 d, 4 f, 6 s\)
(d) \(5 f, 6 d, 7 s, 7 p\)
II. Based upon the above information, solve the questions given below :
(a) Which of the following orbitals has the lowest energy? \(4 d, 4 f, 5 s, 5 p\)
(b) Which of the following orbitals has the highest energy? \(5 p, 5 d, 5 f, 6 s, 6 p\)

Answer: (I) (a) \((n+l)\) values are \(1 s=1+0=1,2 s=2+0=2,3 s=3+0=3\), \(2 p=2+1=3\)
Hence, increasing order of their energy is \(1 s<2 s<2 p<3 s\).
(b) \(4 s=4+0=4,3 s=3+0=3,3 p=3+1=4,4 d=4+2=6\). Hence, \(3 s<3 p<4 s<4 d\).
(c)
\(
\begin{aligned}
& 5 p=5+1=6,4 d=4+2=6,5 d=5+2=7,4 f=4+3=7,6 s=6 \\
& +0=6 .
\end{aligned}
\)
Hence, \(4 d<5 p<6 s<4 f<5 d\).
(d) \(5 f=5+3=8,6 d=6+2=8,7 s=7+0=7,7 p=7+1=8\). Hence, \(7 s<5 f<6 d<7 p\).
(II) (a) \(4 d=4+2=6,4 f=4+3=7,5 s=5+0=5,7 p=7+1=8\). Hence, \(5 s\) has the lowest energy.
(b)
\(
\begin{aligned}
& 5 p=5+1=6,5 d=5+2=7,5 f=5+3=8,6 s=6+0=6,6 p=6 \\
& +1=7
\end{aligned}
\)
Hence, \(5 f\) has the highest energy.

Q69. Which of the following will not show deflection from the path on passing through an electric field?
Proton, cathode rays, electron, neutron.

Answer: Neutrons being neutral will not show deflection from the path on passing through an electric field.
Proton, cathode rays and electrons being the charged particles will show deflection from the path on passing through an electric field.

Q70. Wavelengths of different radiations are given below:
\(
\lambda(\mathrm{A})=300 \mathrm{~nm} \quad \lambda(\mathrm{B})=300 \mu \mathrm{m} \quad \lambda(\mathrm{C})=3 \mathrm{~nm} \quad \lambda(\mathrm{D})=30 Å
\)
Arrange these radiations in the increasing order of their energies.

Answer:

\(
\begin{aligned}
& E=h f \text { or }=\frac{h c}{\lambda} \text { or } E \propto \frac{1}{\lambda} \\
& \lambda(A)=300 \mathrm{~nm}=300 \times 10^{-9} \mathrm{~m} \text { or }=3 \times 10^{-7} \mathrm{~m} ; \\
& \lambda(B)=300 \times 10^{-6} \mathrm{~m}=3 \times 10^{-4} \mathrm{~m} \\
& \lambda(C)=3 \times 10^{-9} \mathrm{~m}, \lambda(D)=30 \times 10^{-10} \mathrm{~m}=3 \times 10^{-9} \mathrm{~m}
\end{aligned}
\)
\(\therefore\) Increasing order of energies is:
\(
B<A<C=D
\)

Q71. The electronic configuration of valence shell of \(\mathrm{Cu}\) is \(3 d^{10} 4 s^1\) and not \(3 d^9 4 s^2\). How is this configuration explained?

Answer: Configuration with completely filled and half-filled orbitals have extra stability. \(\ln 3 \mathrm{~d}^{10} 4 \mathrm{~s}^1\), d-orbitals are completely filled and s-orbital is half-filled. Hence, it is a more stable configuration.

Q72. The Balmer series in the hydrogen spectrum corresponds to the transition from \(n_1=2\) to \(n_2=3,4, \ldots \ldots \ldots\). This series lies in the visible region. Calculate the wave number of line associated with the transition in the Balmer series when the electron moves to \(n=4\) orbit. \(\left(\mathrm{R}_{\mathrm{H}}=109677 \mathrm{~cm}^{-1}\right)\)

Answer:

\(
\begin{aligned}
\bar{\nu} =\mathrm{R}_{\mathrm{H}}\left(\frac{1}{n_1^2} \cdot-\frac{1}{n_2^2}\right) \mathrm{cm}^{-1} \\
 =109677\left(\frac{1}{2^2}-\frac{1}{4^2}\right)=109677\left(\frac{1}{4}-\frac{1}{16}\right) \\
 =20564.44 \mathrm{~cm}^{-1}
\end{aligned}
\)

Q73. According to de Broglie, matter should exhibit dual behaviour, that is both particle and wave-like properties. However, a cricket ball of mass \(100 \mathrm{~g}\) does not move like a wave when it is thrown by a bowler at a speed of \(100 \mathrm{~km} / \mathrm{h}\). Calculate the wavelength of the ball and explain why it does not show wave nature.

Answer:

\(
\begin{aligned}
& m=100 \mathrm{~g} \text { or } 0.1 \mathrm{~kg} \\
& \qquad v=100 \mathrm{~km} / \mathrm{h}=\frac{100 \times 1000}{60 \times 60}=\frac{1000}{36} \mathrm{~m} / \mathrm{s}
\end{aligned}
\)
\(
\lambda=\frac{h}{m v}=\frac{6.63 \times 10^{-34} \mathrm{~kg} \mathrm{~m}^2 \mathrm{~s}^{-1}}{(0.1 \mathrm{~kg})\left(1000 / 36 \mathrm{~ms}^{-1}\right)}=2.387 \times 10^{-34} \mathrm{~m}
\)
Since the wavelength is very small, the wave nature cannot be detected.

Q74. What is the experimental evidence in support of the idea that electronic energies in an atom are quantized?

Answer: The line spectrum of any element has lines corresponding to definite wavelengths. Lines are obtained as a result of electronic transitions between the energy levels. Hence, the electrons in these levels have fixed energy, i.e., quantized values.

Q75. Out of electron and proton which one will have, a higher velocity to produce matter waves of the same wavelength? Explain it.

Answer: Being lighter particles, electrons will have higher velocity.
Hint : \(\lambda=\frac{h}{\mathrm{mv}}\)
From de-Broglie equation, wavelength, \(\lambda=\frac{\mathbf{h}}{\mathrm{mv}}\)
For the same wavelength for two different particles, i.e., electron and proton, \(\mathrm{m}_1 \mathrm{v}_1=\mathrm{m}_2 \mathrm{v}_2\) ( \(\mathrm{h}\) is constant). The lesser the mass of the particle, the greater will be the velocity. Hence, electrons will have higher velocity.

Q76. A hypothetical electromagnetic wave is shown in Fig. 2.2. Find out the wavelength of the radiation.

Answer: Wavelength is the distance between two successive peaks or two successive troughs of a wave. So \(\lambda=4 \times 2.16 \mathrm{pm}=8.64 \mathrm{pm}\)
\(
=8.64 \times 10^{-12} \mathrm{~m} \quad\left[\because 1 \mathrm{pm}=10^{-12} \mathrm{~m}\right]
\)

Q77. Chlorophyll present in green leaves of plants absorbs light at \(4.620 \times 10^{14} \mathrm{~Hz}\). Calculate the wavelength of radiation in nanometer. Which part of the electromagnetic spectrum does it belong to?

Answer: Wavelength, \(\lambda=\frac{\mathrm{c}}{\mathrm{v}}\)
Given, frequency \(\mathrm{v}=4.620 \times 10^{14} \mathrm{~Hz}\) or \(4.620 \times 10^{14} \mathrm{~s}^{-1}\)
\(
\begin{aligned}
& \lambda=\frac{\mathrm{c}}{\mathrm{v}}=\frac{3 \times 10^8 \mathrm{~ms}^{-1}}{4.620 \times 10^{14} \mathrm{~s}^{-1}} \\
& =0.6494 \times 10^{-6} \mathrm{~m} \\
& =649.4 \mathrm{~nm} \quad\left[\because 1 \mathrm{~nm}=10^{-6} \mathrm{~m}\right]
\end{aligned}
\)
Thus, it belongs to the visible region(visible light).

Q78. What is the difference between the terms orbit and orbital?

Answer: 

\(
\begin{array}{|l|l|}
\hline {\text { Differences between Orbit and Orbitals }} \\
\hline \text { Orbit } & \text { Orbitals } \\
\hline \begin{array}{l}
\text { An orbit is the simple planar representation of an } \\
\text { electron. }
\end{array} & \begin{array}{l}
\text { An orbital refers to the dimensional motion of } \\
\text { an electron around the nucleus in a three- } \\
\text { dimensional motion. }
\end{array} \\
\hline \begin{array}{l}
\text { It can be defined as the path that gets } \\
\text { established in a circular motion by revolving the } \\
\text { electron around the nucleus }
\end{array} & \begin{array}{l}
\text { An orbital can be defined as the space or } \\
\text { region where the electron is most likely to be } \\
\text { found. }
\end{array} \\
\hline \begin{array}{l}
\text { The shape of molecules cannot be explained by } \\
\text { an orbit as they are non-directional by nature. }
\end{array} & \begin{array}{l}
\text { The shapes of the molecules can be found } \\
\text { out as they are directional by nature. }
\end{array} \\
\hline \begin{array}{l}
\text { A well-defined orbit goes against the Heisenberg } \\
\text { principle. }
\end{array} & \begin{array}{l}
\text { An ideal orbital agrees with the theory of } \\
\text { Heisenberg’s Principles. }
\end{array} \\
\hline
\end{array}
\)

Q79. Table-tennis ball has a mass \(10 \mathrm{~g}\) and a speed of \(90 \mathrm{~m} / \mathrm{s}\). If speed can be measured within an accuracy of \(4 \%\) what will be the uncertainty in speed and position?

Answer: Given that, speed \(=90 \mathrm{~m} / \mathrm{s}\)
mass \(=10 \mathrm{~g}=10 \times 10^{-3} \mathrm{~kg}\)
Uncertainty in speed \((\Delta \mathrm{v})=4 \%\) of \(90 \mathrm{~ms}^{-1}=\frac{40 \times 90}{100}=3.6 \mathrm{~ms}^{-1}\)
From the Heisenberg uncertainty principle,
\(
\Delta \mathrm{x} \cdot \Delta \mathrm{v}=\frac{\mathrm{h}}{4 \pi \mathrm{m}}
\)
or \(\Delta \mathrm{x}=\frac{\mathrm{h}}{4 \pi \mathrm{m} \Delta \mathrm{v}}\)
Uncertainty in position,
\(
\begin{aligned}
& \Delta \mathrm{x}=\frac{6.626 \times 10^{-34} \mathrm{~kg} \mathrm{~m}^2 \mathrm{~s}^{-1}}{4 \times 3.14 \times 10 \times 10^{-3} \mathrm{~kg} \times 3.6 \mathrm{~ms}^{-1}} \\
& =1.46 \times 10^{-33} \mathrm{~m}
\end{aligned}
\)

Q80. The effect of the uncertainty principle is significant only for the motion of microscopic particles and is negligible for the macroscopic particles. Justify the statement with the help of a suitable example.

Answer: If the uncertainty principle is applied to an object of mass, say about a milligram \(\left(10^{-6} \mathrm{~kg}\right)\), then
\(
\begin{aligned}
& \Delta \mathrm{v} \cdot \Delta \mathrm{x}=\frac{\mathrm{h}}{4 \pi \mathrm{m}} \\
& \Delta \mathrm{v} \cdot \Delta \mathrm{x}=\frac{6.626 \times 10^{-34} \mathrm{~kg} \mathrm{~m}^2 \mathrm{~s}^{-1}}{4 \times 3.14 \times 10^{-6} \mathrm{~kg}} \\
& =0.52 \times 10^{-28} \mathrm{~m}^2 \mathrm{~s}^{-1}
\end{aligned}
\)
The value of \(\Delta \mathrm{v} \cdot \Delta \mathrm{x}\) obtained is extremely small and is insignificant. Therefore, for milligram-sized or heavier objects, the associated uncertainties are hardly of any real consequence.

Q81. The hydrogen atom has only one electron, so mutual repulsion between electrons is absent. However, in multielectron atoms mutual repulsion between the electrons is significant. How does this affect the energy of an electron in the orbitals of the same principal quantum number in multielectron atoms?

Answer: The energy of the electron is determined by the value of \(n\) in the hydrogen atom and by \(n+l\) in the multielectron atom. So for a given principal quantum number electrons of \(s, p, d\) and \(f\) orbitals have different energy (for \(\mathrm{s}, \mathrm{p}, \mathrm{d}\) and \(\mathrm{f}, l=0,1,2\) and 3 respectively).

Q82. What is the photoelectric effect? State the result of the photoelectric effect experiment that could not be explained on the basis of laws of classical physics. Explain this effect on the basis of the quantum theory of electromagnetic radiation.

Answer: Photoelectric effect When radiation with a certain minimum frequency \(\left(\mathrm{f}_0\right)\) strikes the surface of a metal, the electrons are ejected from the surface of the metal. This phenomenon is called the photoelectric effect. The electrons emitted are called photoelectrons.

Equipment for studying the photoelectric effect. Light of a particular frequency strikes a clean metal surface inside a vacuum chamber.
Electrons are ejected from the metal and are counted by a detector that measures their kinetic energy.
The results observed in this experiment were
(i) The electrons are ejected from the metal surface as soon as the beam of light strikes the surface, i.e., there is no time lag between the striking of the light beam and the ejection of electrons from the metal surface.
(ii) The number of electrons ejected is proportional to the intensity or brightness of light.
(iii) For each metal, there is a characteristic minimum frequency, \(\mathbf{f}_0\) (also known as threshold frequency) below which photoelectric effect is not observed. At a frequency \(\mathbf{f}>\mathbf{f}_0\), the ejected electrons come out with certain kinetic energy. The kinetic energies of these electrons increase with the increase of the frequency of the light used.
The above observation cannot be explained by the electromagnetic wave theory. According to this theory, since radiations were continuous, therefore, it should be possible to accumulate energy on the surface of the metal, irrespective of its frequency and thus, radiations of all frequencies should be able to eject electrons.
Similarly, according to this theory, the energy of the electrons ejected should depend upon the intensity of the incident radiation.
Particle Nature of Electromagnetic Radiation
To explain the phenomena of ‘black body radiation’ and ‘photoelectric effect’, Max Planck in 1900, put forward a theory known after his name as Planck’s quantum theory. This theory was further extended by Einstein in 1905.
The important points of this theory are as follows:
(i) The radiant energy is emitted or absorbed not continuously but discontinuously in the form of small discrete packets of energy. Each such packet of energy is called a ‘quantum’. In the case of light, the quantum of energy is called a ‘photon’.
(ii) The energy of each quantum is directly proportional to the frequency of the radiation, i.e., \(\mathrm{E} \propto \mathrm{v}\) or \(\mathrm{E}=\mathrm{hv}\) where, \(\mathrm{h}\) is a proportionality constant, called Planck’s constant. Its value is approximately equal to \(6.626 \times 10^{-27} \mathrm{erg} \mathrm{s}\) or \(6.626 \times 10^{-34} \mathrm{~J} \mathrm{~s}\). (ii) The total amount of energy emitted or absorbed by a body will be some whole number quanta. Hence, \(\mathrm{E}=\mathrm{nhv}\) (where, \(\mathrm{n}\) is any integer)

Q83. Threshold frequency, \(v_0\) is the minimum frequency which a photon must possess to eject an electron from a metal. It is different for different metals. When a photon of frequency \(1.0 \times 10^{15} \mathrm{~s}^{-1}\) was allowed to hit a metal surface, an electron having \(1.988 \times 10^{-19} \mathrm{~J}\) of kinetic energy was emitted. Calculate the threshold frequency of this metal. Show that an electron will not be emitted if a photon with a wavelength equal to \(600 \mathrm{~nm}\) hits the metal surface.

Answer: 

\(
\text { Hint : } h f=h f_0+\frac{1}{2} \mathrm{mv}^2
\)
Energy of photon \(=\) Threshold energy + Kinetic energy
\(
h f=h f_0+K E
\)
Where, \({f}_0=\) threshold frequency
\(f=\) frequency of the photon striking the surface
According to the question:
Frequency of photon \(f=1 \times 10^{15} \mathrm{~s}^{-1}\)
Kinetic energy \(=1.988 \times 10^{-19} \mathrm{~J}\)
Photoelectric effect: \(h f=h f_0+K E\)
\(
\begin{aligned}
& \left(6.626 \times 10^{-34} \times 1 \times 10^{15}\right) \\
& =\left(6.626 \times 10^{-34} \times f_0\right)+\left(1.988 \times 10^{-19}\right) \mathrm{J} \\
& f_0=\frac{\left(6.626 \times 10^{-19}\right)-\left(1.988 \times 10^{-19}\right)}{6.626 \times 10^{-34}} \\
& f_0=7 \times 10^{14} \mathrm{~Hz}
\end{aligned}
\)
Checking whether photon of \(600 \mathrm{~nm}\) would eject electron or not.
Wavelength of the given photon \(=600 \mathrm{~nm}\)
\(
=600 \times 10^{-9} \mathrm{~m}
\)
Frequency of given photon striking metal surface
\(
\begin{aligned}
& f=\frac{c}{\lambda} \\
& f=\frac{3 \times 10^0}{600 \times 10^{-9}} \\
& =5 \times 10^{14} \mathrm{~Hz} \\
& f=5 \times 10^{14} \mathrm{~Hz}
\end{aligned}
\)
Since \(f<f_0\left(v_0=7 \times 10^{14} \mathrm{~Hz}\right)\)
Hence, photoelectrons will not be ejected.

Q84. When an electric discharge is passed through hydrogen gas, the hydrogen molecules dissociate to produce excited hydrogen atoms. These excited atoms emit electromagnetic radiation of discrete frequencies which can be given by the general formula
\(

(\tilde{\nu})=109677 \left[\frac{1}{n_i^2}-\frac{1}{n_f^2}\right]
\)
What points of Bohr’s model of an atom can be used to arrive at this formula? Based on these points derive the above formula giving a description of each step and each term.

Answer: The two important points of Bohr’s model that can be used to derive the given formula are as follows
(i) Electrons revolve around the nucleus in a circular path of fixed radius and energy. These paths are called orbits, stationary states or allowed energy states.
(ii) Energy is emitted or absorbed when an electron moves from a higher stationary state to a lower stationary state or from a lower stationary state to a higher stationary state to a lower stationary state to a higher stationary state respectively.
Derivation The energy of the electron in the \(\mathrm{n}^{\text {th }}\) stationary state is given by the expression,
\(
\mathrm{E}_{\mathrm{n}}=-\mathrm{R}_{\mathrm{H}}\left(\frac{1}{\mathrm{n}^2}\right) \quad \mathrm{n}=1,2,3 \quad \ldots \ldots \ldots \ldots(\mathrm{i})
\)
Where \(R_H\) is called Rydberg constant and its value is \(2.18 \times 10^{-18} \mathrm{~J}\). The energy of the lowest state, also called the ground state, is \(\mathrm{E}_{\mathrm{n}}=-2.18 \times 10^{-18}\left(\frac{1}{1^2}\right)=-2.18 \times 10^{-18} \mathrm{~J} \ldots \ldots \ldots \ldots\) (ii)
The energy gap between the two orbits is given by the equation,
\(
\Delta \mathrm{E}=\mathrm{E}_{\mathrm{f}}-\mathrm{E}_{\mathrm{i}}
\)
On combining Eqs. (i) and (ii)
\(
\Delta \mathrm{E}=\left(-\frac{\mathrm{R}_{\mathrm{H}}}{\mathrm{n}_{\mathrm{f}}^2}\right)-\left(-\frac{\mathrm{R}_{\mathrm{H}}}{\mathrm{n}_{\mathrm{i}}^2}\right)
\)
Where \(n_i\) and \(n_f\) stand for initial orbit and final orbit.
\(
\Delta \mathrm{E}=\mathrm{R}_{\mathrm{H}}\left[\frac{1}{n_{\mathrm{i}}^2}-\frac{1}{\mathrm{n}_{\mathrm{f}}^2}\right]=2.18 \times 10^{-18} \mathrm{~J}\left[\frac{1}{n_{\mathrm{i}}^2}-\frac{1}{n_{\mathrm{f}}^2}\right]
\)
Frequency, \((\tilde{\nu})\) associated with the absorption and emission of the photon can be calculated as follows
\(
\begin{aligned}
&
(\tilde{\nu})=\frac{\Delta \mathrm{E}}{\mathrm{h}}=\frac{\mathrm{R}_{\mathrm{H}}}{\mathrm{h}}\left[\frac{1}{\mathrm{n}_{\mathrm{i}}^2}-\frac{1}{\mathrm{n}_{\mathrm{f}}^2}\right] \\
& \Rightarrow (\tilde{\nu})=\frac{2.18 \times 10^{-18} \mathrm{~J}}{6.626 \times 10^{-34} \mathrm{Js}}\left[\frac{1}{\mathrm{n}_{\mathrm{i}}^2}-\frac{1}{\mathrm{n}_{\mathrm{f}}^2}\right] \\
& (\tilde{\nu})=3.29 \times 10^{15}\left[\frac{1}{\mathrm{n}_{\mathrm{i}}^2}-\frac{1}{\mathrm{n}_{\mathrm{f}}^2}\right] \mathrm{Hz} \\
& \Rightarrow
(\tilde{\nu})=\frac{\mathrm{v}}{\mathrm{c}}=\frac{329 \times 10^{15}}{3 \times 10^8 \mathrm{~ms}^{-1}}\left[\frac{1}{\mathrm{n}_{\mathrm{i}}^2}-\frac{1}{\mathrm{n}_{\mathrm{f}}^2}\right] \\
&
(\tilde{\nu})=1.09577 \times 10^7\left[\frac{1}{\mathrm{n}_{\mathrm{i}}^2}-\frac{1}{\mathrm{n}_{\mathrm{f}}^2}\right] \mathrm{m}^{-1} \\
&
(\tilde{\nu})=109677\left[\frac{1}{\mathrm{n}_{\mathrm{i}}^2}-\frac{1}{\mathrm{n}_{\mathrm{f}}^2}\right] \mathrm{cm}^{-1}
\end{aligned}
\)

Q85. Calculate the energy and frequency of the radiation emitted when an electron jumps from \(n=3\) to \(n=2\) in a hydrogen atom.

Answer: In the hydrogen spectrum, the spectral lines are expressed in terms of wave number \((\tilde{\nu})\) obey the following formula Wave number, \((\tilde{\nu})=\mathrm{R}_{\mathrm{H}}\left(\frac{1}{\mathrm{n}_1^2}-\frac{1}{\mathrm{n}_2^2}\right) \quad\left(\right.\) where, \(\mathrm{R}_{\mathrm{H}}=\) Rydberg constant \(\left. 109677 \mathrm{~cm}^{-1}\right)\)
\(
\begin{aligned}
& (\tilde{\nu})=109677 \mathrm{~cm}^{-1}\left(\frac{1}{2^2}-\frac{1}{3^2}\right) \\
& (\tilde{\nu})=108677 \times \frac{5}{36}=15232.9 \mathrm{~cm}^{-1} \\
& (\tilde{\nu})=\frac{1}{\lambda} \\
& \text { or, } \lambda=\frac{1}{(\tilde{\nu})}=\frac{1}{15232.9}=6.564 \times 10^{-5} \mathrm{~cm} \\
& \text { Wavelength, } \lambda=6.564 \times 10^{-7} \mathrm{~m} \\
& \text { Energy, E }=\frac{\mathrm{hc}}{\lambda} \\
& =\frac{6.626 \times 10^{-34} \mathrm{Js} \times 3.0 \times 10^8 \mathrm{~ms}^{-1}}{6.564 \times 10^{-7} \mathrm{~m}} \\
& =3.028 \times 10^{-19} \mathrm{~J} \\
& \text { Frequency, } f=\frac{\mathrm{c}}{\lambda}=\frac{3.0 \times 10^8 \mathrm{~m}^{-1}}{6.564 \times 10^{-7} \mathrm{~m}} \\
& =-0.457 \times 10^{15} \mathrm{~s}^{-1}=4.57 \times 10^{14} \mathrm{~s}^{-1}
\end{aligned}
\)

Q86. Why was a change in the Bohr Model of the atom required? Due to which important development (s), the concept of movement of an electron in an orbit was replaced by, the concept of probability of finding an electron in an orbital? What is the name given to the changed model of the atom?

Answer: In the Bohr model, an electron is regarded as a charged particle moving in well-defined circular orbits about the nucleus. An orbit can completely be defined only if both the position and the velocity of the electron are known exactly at the same time. This is not possible according to the Heisenberg uncertainty principle. Furthermore, the wave character of the electron is not considered in the Bohr model. Therefore, the concept of the movement of an electron in an orbit was replaced by the concept of probability of finding an electron in an orbital due to the de-Broglie concept of the dual nature of the electron and Heisenberg’s uncertainty principle. The changed model is called the quantum mechanical model of the atom.