NEET Practice Q & A

Hint

\(
\text { (b) } \Delta L=\frac{n h}{2 \pi}-\frac{(n-1) h}{2 \pi}=\frac{h}{2 \pi}
\)

Hint

(d) Orbital angular momentum of the electron.
Explanation:
In a hydrogen-like atom, the orbital angular momentum is quantized and given by the formula \(L=n h / 2 \pi\), where n is the principal quantum number. Since the ground state corresponds to \(\)n=1\(\), the orbital angular momentum is the same for all hydrogen-like atoms and ions in their ground states regardless of their atomic number.

Hint

(d) Fifth orbit to third orbit. The second line in the Paschen series occurs when an electron transitions from the fifth orbit ( \(n=5\) ) to the third orbit ( \(n=3\) ). The Paschen series is characterized by transitions to the \(\mathrm{n}=3\) energy level. The first line corresponds to a transition from \(n=4\) to \(n=3\), the second line from \(n=5\) to \(n=3\), and so on.

Hint

\(
\text { (d) Impact parameter, } b \propto \cot \frac{\theta}{2} \Rightarrow \frac{b_1}{b_2}=\frac{\cot 45^{\circ}}{\cot 60^{\circ}}=\sqrt{3}
\)

Hint

(c) The minimum energy required to ionise an atom is equal to the energy to excite an electron in the atom from its ground state corresponding to \(n=1\), to its ionisation level corresponding to \(n=\infty\). If the amount of energy given to atom is equal to the ionisation energy, then the electron is just able to become a free electron and the atom is about to be ionised.
When this occurs, one outermost electron from the atom is just to be removed from the atom. Much higher energy is necessary to free an innermost electron from the atom because this will occur only after the outermost electrons are all removed.

Hint

(b) \(v \propto \frac{Z}{n}, \quad\) since \(Z\) and \(n\) both have become twice. Thus, velocity of electron in second orbit of \(\mathrm{He}^{+}\)will be \(v\).

Hint

\(
\text { (d) If } n=4 \text {, then spectral lines emitted }=\frac{n(n-1)}{2}=6
\)

Hint

\(
\begin{aligned}
&\text { (a) } L=\frac{n h}{2 \pi} \text {, it does not depend on } Z \text {. }\\
&\therefore n_2>n_1
\end{aligned}
\)

Hint

(c) \(\frac{1}{\lambda}=R\left(\frac{1}{2^2}-\frac{1}{4^2}\right)=R\left(\frac{16-4}{16 \times 4}\right)=\frac{12 R}{16 \times 4}=\frac{3}{16} R\)
or \(\lambda=\frac{16}{3} R\)
\(
\begin{aligned}
{\nu} & =\frac{c}{\lambda}=\frac{3 \times 10^8 \times 10^7 \times 3}{16}\left(\because R=10^5 \mathrm{~cm}^{-1},=10^7 \mathrm{~m}^{-1} \text { given }\right) \\
& =\frac{9}{16} \times 10^{15} \mathrm{~Hz}
\end{aligned}
\)

Hint

\(
\begin{aligned}
&\text { (c) Wave number is reciprocal of wavelength, }\\
&\begin{aligned}
\bar{\nu}=k =\frac{1}{\lambda}=R\left[\frac{1}{n_1^2}-\frac{1}{n_2^2}\right]=R\left[\frac{1}{2^2}-\frac{1}{4^2}\right] \\
=R\left[\frac{1}{4}-\frac{1}{16}\right]=\frac{3 R}{16}
\end{aligned}
\end{aligned}
\)

Hint

(a) Use Bragg’s X-ray diffraction law,
\(
\begin{aligned}
& n \lambda & =2 d \sin \theta \\
\therefore & \lambda & =\frac{2 d \sin \theta}{n}
\end{aligned}
\)
For longest wavelength take \(\sin \theta=1\)
\(
\begin{aligned}
\therefore \quad \lambda & =\frac{2 \times 2.82 \times 1}{2} \quad(\because n=2, \text { for second order }) \\
& =2.82 Å
\end{aligned}
\)

Hint

(b) According to the question,
For the distance of closest approach kinetic energy will totally be converted to potential energy.
Hence, \(\frac{1}{2} m v^2=\frac{1}{4 \pi \varepsilon_0} \frac{Q q}{r_0} \Rightarrow r_0=\frac{Q q}{2 \pi \varepsilon_0 m v^2}\)

Hint

(a) \(E \propto m \quad\) and \(\quad r \propto \frac{1}{m}\) i.e. energy will become two times and radius will become half.
Hence, \(E_0=2(-13.6)=-27.2 \mathrm{eV}\) and \(r_0=a_0 / 2\)

Hint

\(
\begin{array}{ll}
\text { (b) } & r_3=(3)^2 \cdot r_1=9 x \\
\text { Now, } & 3 \lambda_3=\left(2 \pi r_3\right)=18 \pi x \\
\therefore & \lambda_3=6 \pi x
\end{array}
\)

Hint

(a) According to scattering formula,
\(
\begin{array}{ll}
& N \propto \frac{1}{\sin ^4(\theta / 2)} \Rightarrow \frac{N_2}{N_1}=\left[\frac{\sin \left(\theta_1 / 2\right)}{\sin \left(\theta_2 / 2\right)}\right]^4 \\
\Rightarrow & \frac{N_2}{N_1}=\left[\frac{\sin \frac{90^{\circ}}{2}}{\sin \frac{60^{\circ}}{2}}\right]^4=\left[\frac{\sin 45^{\circ}}{\sin 30^{\circ}}\right]^4 \\
\Rightarrow & N_2=(\sqrt{2})^4 \times N_1=4 \times 56=224
\end{array}
\)

Hint

\(
\text { (d) } \alpha \text {-particles cannot be attracted by the nucleus. }
\)

Hint

(d) \(L_n=\frac{n h}{2 \pi} \Rightarrow I \omega=\frac{n h}{2 \pi} \Rightarrow\left(m r^2\right) \omega=\frac{n h}{2 \pi}\)
As, \(\quad r \propto n^2\)
\(\therefore \quad \omega \propto \frac{1}{n^3}\)

Hint

(d) \(\Delta E \propto\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)\)
Also, \(\Delta E \propto \frac{1}{\lambda}\)
Since, \(\lambda\) is less for transition from \(n=5\) to \(n=2[latex] than for [latex]n=4\) to \(n=2\).
From the given option, only \(\lambda_{\text {violet }}<\lambda_{\text {blue }}\). Thus, light of violet colour would have been emitted.

Hint

\(
\begin{aligned}
&\text { (c) Change in the angular momentum, }\\
&\begin{gathered}
\Delta L=L_{n_2}-L_{n_1}=\frac{n_2 h}{2 \pi}-\frac{n_1 h}{2 \pi} \\
\Delta L=\frac{h}{2 \pi}\left(n_2-n_1\right)=\frac{6.6 \times 10^{-34}}{2 \times 3.14}(5-4)=1.05 \times 10^{-34} \mathrm{~J}-\mathrm{s}
\end{gathered}
\end{aligned}
\)

Hint

(b) In a hydrogen atom the time period is given by \(T \propto n^3\)
\(
\frac{T_1}{T_2}=\left(\frac{n_1}{n_2}\right)^3 \Rightarrow \frac{8}{1}=\left(\frac{n_1}{n_2}\right)^3 \Rightarrow \frac{n_1}{n_2}=\frac{2}{1}
\)
Thus, the values must be \(n_1=2\) and \(n_2=1\).

Hint

\(
\begin{aligned}
&\text { (d) According to Moseley’s law, } \sqrt{v} \propto Z\\
&\therefore \quad \frac{\mathrm{v}_A}{\mathrm{v}_B}=\left(\frac{Z_A}{Z_B}\right)^2=\left(\frac{1}{2}\right)^2=\frac{1}{4}
\end{aligned}
\)

Hint

\(
\begin{array}{rlrl}
\text { (c) } & Z^2\left[\frac{13.6}{4}-\frac{13.6}{9}\right] & =47.2 \\
& \therefore Z & =5
\end{array}
\)

Hint

\(
\begin{aligned}
&\begin{aligned}
& \text { (a) Excitation energy, } \Delta E=E_2-E_1=13.6 Z^2\left[\frac{1}{1^2}-\frac{1}{2^2}\right] \\
& \therefore \quad 40.8=13.6 Z^2 \times \frac{3}{4} \\
& \therefore \quad Z=2
\end{aligned}\\
&\text { So, required energy to remove the electron from ground state }\\
&=+\frac{13.6 Z^2}{(1)^2}=13.6(2)^2=54.4 \mathrm{eV}
\end{aligned}
\)

Hint

(c) Final energy of electron \(=-13.6+121=-1.51 \mathrm{eV}\)
This energy corresponds to third level, i.e. \(n=3\) Hence, number of spectral lines emitted
\(
=\frac{n(n-1)}{2}=\frac{3(3-1)}{2}=3
\)

Hint

\(
\begin{aligned}
& \text { (c) As, } m v r=\frac{h}{2 \pi} \text { (in first orbit) } \\
& \therefore \quad v=\frac{h}{2 \pi m r} \Rightarrow a=\frac{v^2}{r}=\frac{h^2}{4 \pi^2 m^2 r^3}
\end{aligned}
\)

Hint

\(
\begin{aligned}
& \text { (c) } \frac{1}{\lambda}=R Z^2\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right) \Rightarrow \lambda \propto \frac{1}{Z^2} \\
& \lambda_{\mathrm{Li}^{++}}: \lambda_{\mathrm{He}^{+}}: \lambda_{\mathrm{H}}=\frac{1}{3^2}: \frac{1}{2^2}: \frac{1}{1^2}=\frac{1}{9}: \frac{1}{4}: \frac{1}{1}=4: 9: 36
\end{aligned}
\)

Hint

\(
\text { (b) We know that, } \frac{1}{\lambda_{K_\alpha}}=R(Z-1)^2\left(\frac{1}{1^2}-\frac{1}{2^2}\right)
\)
\(
\begin{aligned}
(Z-1)^2 & =\frac{4}{3} \times \frac{1}{\lambda_{K_\alpha} R}=680 \\
(Z-1) & =26 \quad\left(\because 680 \approx 26^2\right) \\
Z & =27 \rightarrow \text { Cobalt }
\end{aligned}
\)

Hint

(c) For \(K_\alpha\) line, \(\lambda_{K_\alpha} \propto \frac{1}{(Z-1)^2}\)
So, \(\frac{\lambda_2}{\lambda_1}=\left(\frac{Z_1-1}{Z_2-1}\right)^2 \left(\because \lambda_1=\lambda\right)\)
\(
\begin{aligned}
\frac{\lambda_2}{\lambda} & =\left(\frac{43-1}{29-1}\right)^2=\left(\frac{42}{28}\right)^2 \\
\lambda_2 & =\frac{9}{4} \lambda
\end{aligned}
\)

Hint

(a) Magnetic moment \(\mu\) of a revolving electron is given by
\(
\mu=\frac{e}{2 m} L=\frac{e}{2 m} \cdot \frac{n h}{2 \pi}=\left(\frac{e h}{2 m}\right) \cdot \frac{n}{2 \pi}=\frac{n}{2 \pi} \mu_B
\)
where, \(\quad \mu_B=\frac{e h}{2 m}\) is Bohr magneton.
\(\mu \propto n\)

Hint

\(
\begin{aligned}
& \text { (c) } a=\frac{v^2}{r} \text { or } a \propto \frac{\left(Z^2\right)}{(1 / Z)} \\
& \therefore \quad a \propto Z^3
\end{aligned}
\)
For singly ionised helium atom, \(Z=2\) and doubly ionised lithium atom, \(Z=3\)
\(
\Rightarrow \quad \frac{a_{\mathrm{He}^{+}}}{a_{\mathrm{Li}^{++}}}=\frac{(2)^3}{(3)^3}=\frac{8}{27}
\)

Hint

(b) For transition \(n=3\) to \(n=2\),
\(
\frac{1}{\lambda_1}=R\left[\frac{1}{2^2}-\frac{1}{3^2}\right]=\frac{5 R}{36} \dots(i)
\)
For transition \(n=4\) to \(n=2\),
\(
\frac{1}{\lambda_2}=R\left[\frac{1}{2^2}-\frac{1}{4^2}\right]=\frac{3 R}{16} \dots(ii)
\)
Now, dividing Eq. (i) by Eq. (ii), we get
\(
\frac{\lambda_2}{\lambda_1}=\frac{20}{27} \text { or } \lambda_2=\frac{20}{27} \lambda_1=\frac{20}{27} \lambda_0 \quad\left(\because \lambda_1=\lambda_0\right)
\)

Hint

\(
\begin{aligned}
&\text { (a) Wavelength is given by }\\
&\begin{array}{ll}
& \frac{1}{\lambda}=R\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right) \\
\Rightarrow & \frac{1}{\lambda}=R\left[\frac{1}{(2)^2}-\frac{1}{(4)^2}\right] \\
\Rightarrow & \frac{1}{\lambda}=R\left(\frac{1}{4}-\frac{1}{16}\right) \Rightarrow \frac{1}{\lambda}=R\left(\frac{4-1}{16}\right) \\
\Rightarrow & \frac{1}{\lambda}=\frac{3 R}{16} \Rightarrow \lambda=\frac{16}{3 R}
\end{array}
\end{aligned}
\)

Hint

(a) We know that, frequency, \(\nu=R c\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)\) This gives,
\(
\begin{aligned}
& {\nu}_1=R c\left(1-\frac{1}{\infty}\right)=R c \\
& \nu_2=R c\left(1-\frac{1}{4}\right)=\frac{3}{4} R c \\
& \nu_3=R c\left(\frac{1}{4}-\frac{1}{\infty}\right)=\frac{R c}{4} \Rightarrow {\nu}_1-{\nu}_2={\nu}_3
\end{aligned}
\)

Hint

(a) According to the question,
Energy, \(E=-3.4 \mathrm{eV}\)
\(
\begin{array}{lc}
\Rightarrow & -\frac{13.6}{n^2}=-3.4 \\
\Rightarrow & n^2=4 \\
\Rightarrow & n=2
\end{array}
\)
So, the angular momentum of the electron, \(m v r=\frac{n h}{2 \pi}\)
\(
\begin{aligned}
L & =m v r=\frac{2 \times 6.626 \times 10^{-34}}{2 \times 3.14} \\
& =\frac{6.626 \times 10^{-34}}{3.14}=2.1 \times 10^{-34} \mathrm{~J}-\mathrm{s}
\end{aligned}
\)

Hint

(a) The wavelength can be given as
\(
\frac{1}{\lambda}=Z^2 R\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right) \dots(i)
\)
For \(\lambda\) to be shortest, \(\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)\) should be maximum.
So, \(n_2=\infty\) and \(n_1=1\)
Now, from Eq. (i), we get
\(
\frac{1}{\lambda}=R\left(\frac{1}{1}\right) \Rightarrow \lambda=\frac{1}{R}=\frac{1}{10^7} \mathrm{~m} \simeq 1000 Å
\)

Hint

(c) The energy of first excitation of sodium is \(E=h v=\frac{h c}{\lambda}\)
\(
\begin{array}{ll}
\therefore \quad E & =\frac{6.63 \times 10^{-34} \times 3 \times 10^8}{5896 \times 10^{-10}} \mathrm{~J} \\
E & =3.37 \times 10^{-19} \mathrm{~J}
\end{array}
\)
Also, since \(1.6 \times 10^{-19} \mathrm{~J}=1 \mathrm{eV}\)
\(
\begin{array}{ll}
\therefore & E=\frac{3.37 \times 10^{-19}}{1.6 \times 10^{-19}} \\
\Rightarrow & E=2.1 \mathrm{eV}
\end{array}
\)
Hence, corresponding first excitation potential is 2.1 V.

Hint

(a) The wavelength of lines is given by
\(
\frac{1}{\lambda}=R\left(\frac{1}{1^2}-\frac{1}{n^2}\right)
\)
For Lyman series, the shortest wavelength is for \(n=\infty\) and longest is for \(n=2\).
\(
\therefore \begin{aligned}
\frac{1}{\lambda_S} & =R\left(\frac{1}{1^2}\right) \dots(i) \\
\frac{1}{\lambda_L} & =R\left(\frac{1}{1^2}-\frac{1}{2^2}\right)=\frac{3}{4} R \dots(ii)
\end{aligned}
\)
On dividing Eq. (i) by Eq. (ii), we get
\(
\frac{\lambda_L}{\lambda_S}=\frac{4}{3}
\)
Given,
\(
\lambda_S=91.2 \mathrm{~nm} \Rightarrow \lambda_L=91.2 \times \frac{4}{3}=121.6 \mathrm{~nm}
\)

Hint

(d) After removing one electron from helium atom it will become hydrogen like atom.
\(
\therefore \quad E=24.6+(13.6)(2)^2=79 \mathrm{eV}
\)

Hint

\(
\begin{aligned}
& \text { (b) As, } & \frac{1}{\lambda} & =R\left(\frac{1}{1^2}-\frac{1}{n^2}\right) \\
& \therefore & \frac{R}{n^2} & =R-\frac{1}{\lambda}=\frac{\lambda R-1}{\lambda} \\
& \text { or } & n & =\sqrt{\frac{\lambda R}{\lambda R-1}}
\end{aligned}
\)

Hint

\(
\begin{aligned}
& \text { (a) As, } \sqrt{f} \propto(Z-1) \sqrt{\frac{1}{n_1^2}-\frac{1}{n_2^2}} \\
& \therefore \quad \text { Slope } \propto \sqrt{\frac{1}{n_1^2}-\frac{1}{n_2^2}} \\
& \therefore \quad \frac{(\text { Slope })_{K_\beta}}{(\text { Slope })_{K_\alpha}}=\frac{\sqrt{1-\frac{1}{9}}}{\sqrt{1-\frac{1}{4}}}=\sqrt{\frac{8}{9} \times \frac{4}{3}}=\sqrt{\frac{32}{27}}
\end{aligned}
\)

Hint

\(
\text { (a) } \lambda=\frac{h}{p}=\frac{h}{m v} \text { or } \lambda \propto \frac{1}{v} \quad\left(\because v \propto \frac{1}{n}\right)
\)
\(
\therefore \quad \lambda_n \propto n \quad J_n=n \frac{h}{2 \pi},
\)
i.e. \(J_n \propto n\)
Hence, \(J_n \propto \lambda_n\)

Hint

(b) \(\Delta E_{4 \rightarrow 1}=\left(\frac{13.6}{1^2}-\frac{13.6}{16}\right) \mathrm{eV}=12.75 \mathrm{eV}\)
Momentum of hydrogen atom \(=\) Momentum of photon
\(
=\frac{E}{c}=\frac{12.75 \times 1.6 \times 10^{-19}}{3.0 \times 10^8}=6.8 \times 10^{-27} \mathrm{~kg}-\mathrm{ms}^{-1}
\)

Hint

\(
\begin{aligned}
&\text { (b) Since, } r \propto n^2 \text { i.e. } \frac{r_f}{r_i}=\left(\frac{n_f}{n_i}\right)^2\\
&\Rightarrow \frac{21.2 \times 10^{-11}}{5.3 \times 10^{-11}}=\left(\frac{n}{1}\right)^2 \Rightarrow n^2=4 \Rightarrow n=2
\end{aligned}
\)

Hint

(a) For \(\boldsymbol{\alpha}\)-particle, at distance of closest approach, then
Decrease in \(\mathrm{KE}=\) Increase in PE
\(
\begin{array}{rlrl}
2 \mathrm{eV} & =\frac{1}{4 \pi \varepsilon_0} \cdot \frac{2 e \cdot Z e}{r^2} \\
\Rightarrow \quad & Z e & =4 \pi \varepsilon_0 r^2 \mathrm{~V} \dots(i)
\end{array}
\)
For proton, if \(r_1\) be the distance of closest approach Decrease in KE = Increase in PE
\(
\begin{aligned}
R_1^2 & =\frac{1}{4 \pi \varepsilon_0 V}(Z e) \\
& =\frac{1}{4 \pi \varepsilon_0 V} \cdot 4 \pi \varepsilon_0 r^2 V \text { [from Eq. (i)] }\\
\Rightarrow \quad r_1^2 & =r^2 \\
\Rightarrow \quad r_1 & =r
\end{aligned}
\)

Hint

(b) The energy of the projectile will be equivalent to potential energy of the charge system,
\(
E=\frac{1}{4 \pi \varepsilon_0} \frac{Z_1 Z_2}{r_0} \Rightarrow \text { Energy, } E \propto Z_1 Z_2
\)

Hint

(a)

\(
\begin{aligned}
&\text { Centripetal force = force of attraction of nucleus on electron }\\
&\begin{aligned}
& \frac{m v^2}{a_0}=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{e^2}{a_0^2} \\
& v=\frac{e}{\sqrt{4 \pi \varepsilon_0 m a_0}}
\end{aligned}
\end{aligned}
\)

Hint

(b) For the wavelength of Balmer’s series,
\(
\frac{1}{\lambda}=R\left(\frac{1}{2^2}-\frac{1}{n^2}\right)
\)
where, \(R\) is Rydberg’s constant.
For Balmer series, \(n=3\) gives the first member of series and \(n=4\) gives the second member of series. Hence,
\(
\begin{aligned}
& \frac{1}{\lambda_1}=R\left(\frac{1}{2^2}-\frac{1}{3^2}\right)=R\left(\frac{5}{36}\right) \dots(i) \\
& \frac{1}{\lambda_2}=R\left(\frac{1}{2^2}-\frac{1}{4^2}\right)=R\left(\frac{12}{16 \times 4}\right)=\frac{3 R}{16} \dots(ii) \\
& \frac{\lambda_2}{\lambda_1}=\frac{16}{3} \times \frac{5}{36}=\frac{20}{27} \\
& \lambda_1=\lambda \Rightarrow \lambda_2=\frac{20}{27} \lambda
\end{aligned}
\)

Hint

\(
\begin{aligned}
&\text { (a) }\\
&\begin{aligned}
& \text { Here, } E_1=\frac{m e^4}{8 \varepsilon_0^2 h^2} \Rightarrow m e^4=8 \varepsilon_0^2 h^2 E_1 \\
& f_n=\frac{v_n}{2 \pi r_n}=\frac{\left(e^2 / 2 \varepsilon_0 n h\right)}{(2 \pi) \varepsilon_0 n^2 h^2 / \pi m e^2} \\
& \quad=\frac{m e^4}{4 \varepsilon_0^2 n^3 h^3}=\frac{8 E_1 \varepsilon_0^2 h^2}{4 \varepsilon_0^2 n^3 h^3}=\frac{2 E_1}{n^3 h}
\end{aligned}
\end{aligned}
\)

Hint

\(
\begin{aligned}
& \text { (d) As, } a=v^2 / r \\
& r \propto n^2 \text { and } v \propto \frac{1}{n} \\
& \therefore \quad a \propto \frac{1}{n^4} \\
& \Rightarrow \quad \frac{a_M}{a_L}=\left(\frac{2}{3}\right)^4=\frac{16}{81}
\end{aligned}
\)

Hint

(a) Minimum 10.2 eV energy is required to excite a hydrogen atom. In perfectly inelastic collision maximum
\(
\begin{aligned}
\therefore \quad \text { Energy lost } & =\frac{1}{2}(2 m)\left(\frac{v}{2}\right)^2=10.2 \times 1.6 \times 10^{-19} \\
\frac{1}{4} m v^2 & =1.632 \times 10^{-18} \\
\therefore \quad \text { Velocity, } v & =\sqrt{\frac{1.632 \times 4 \times 10^{-18}}{1.67 \times 10^{-27}}}=6.25 \times 10^4 \mathrm{~ms}^{-1}
\end{aligned}
\)

Hint

(d) The ground state of hydrogen ( \(n=1\) ) is represented by \(K\), the first excited state ( \(n=2\) ) is represented by \(L\), the second excited state ( \(n=3\) ) is represent by \(M\). So, the energy transferred by the second electron can be given by
\(
E=-13.6\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)=-13.6\left(\frac{1}{3^2}-\frac{1}{\infty^2}\right) \simeq-1.51 \mathrm{eV}
\)

Hint

\(
\begin{aligned}
&\text { (b) According to Bohr’s theory, the wavelength of the radiation }\\
&\begin{aligned}
\frac{1}{\lambda} & =Z^2 R\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right) \Rightarrow \frac{\nu}{c}=Z^2 R\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right) \\
\nu & =c Z^2 R\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right) \Rightarrow 2.7 \times 10^{15}=c Z^2 R\left(\frac{1}{1^2}-\frac{1}{2^2}\right) \\
\nu & =c Z^2 R\left(\frac{1}{1^2}-\frac{1}{3^2}\right) \Rightarrow v=\frac{32}{27} \times 2.7 \times 10^{15} \\
& =3.2 \times 10^{15} \mathrm{~Hz}
\end{aligned}
\end{aligned}
\)

Hint

(b) Because atom is hollow and whole mass of atom is concentrated in a small centre called nucleus. So, \(A^{\prime}\) will be maximum while \(B^{\prime}\) minimum.

Hint

(d) If \(E\) is the energy radiated in transition, then from figure
\(
E_{R \rightarrow G}>E_{Q \rightarrow S}>E_{R \rightarrow S}>E_{Q \rightarrow R}>E_{P \rightarrow Q}
\)
For obtaining blue line, energy radiated should be more. 

Hint

(d) The transition marked as \(A\) represents series limit of Lyman series, \(B\) represents third member of Balmer series and \(C\) represents second member of Paschen series.

Hint

(c) \(E \propto m\)
\(
\therefore \quad \lambda \propto \frac{1}{m}
\)
In case of hydrogen atom for longest wavelength of photon in the Balmer series,
\(
\frac{1}{\lambda}=R\left(\frac{1}{4}-\frac{1}{9}\right) \text { or } \lambda=\frac{36}{5 R}
\)
\(
\begin{aligned}
&\text { For this atom, } \quad \lambda^{\prime}=\frac{\lambda}{2}=\frac{18}{5 R}\\
&(\because m=2 m)
\end{aligned}
\)

Hint

(b) Let energy corresponding to state \(A, B\) and \(C\) be \(E_A, E_B\) and \(E_C\), respectively.
So, from energy level diagram, or
\(
\begin{aligned}
& \left(E_C-E_B\right)+\left(E_B-E_A\right)=\left(E_C-E_A\right) \\
& \frac{h c}{\lambda_1}+\frac{h c}{\lambda_2}=\frac{h c}{\lambda_3} \Rightarrow \lambda_3=\frac{\lambda_1 \lambda_2}{\lambda_1+\lambda_2}
\end{aligned}
\)

Hint

(a) For transition from \(4 E\) to \(E\),
\(
(4 E-E)=\frac{h c}{\lambda_1} \Rightarrow \lambda_1=\frac{h c}{3 E} \dots(i)
\)
For transition from \(\frac{7}{3} E\) to \(E\),
\(
\left(\frac{7}{3} E-E\right)=\frac{h c}{\lambda_2} \Rightarrow \lambda_2=\frac{3 h c}{4 E} \dots(ii)
\)
From Eqs. (i) and (ii), we get
\(
\frac{\lambda_1}{\lambda_2}=\frac{4}{9}
\)

Hint

(a) Energy of emitted photon, \(E=R h c\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)\)
\(
\begin{array}{ll}
\Rightarrow & R_{(4 \rightarrow 3)}=R h c\left(\frac{1}{3^2}-\frac{1}{4^2}\right)=R h c\left(\frac{7}{9 \times 16}\right)=0.05 R h c \\
& E_{(4 \rightarrow 2)}=R h c\left(\frac{1}{2^2}-\frac{1}{4^2}\right)=R h c\left(\frac{3}{16}\right)=0.2 R h c \\
& E_{(2 \rightarrow 1)}=R h c\left[\frac{1}{(1)^2}-\frac{1}{(2)^2}\right]=R h c\left(\frac{3}{4}\right)=0.75 R h c \\
\text { and } & E_{(1 \rightarrow 4)}=R h c\left[\frac{1}{(4)^2}-\frac{1}{(1)^2}\right]=-\frac{8}{9} R h c=-0.9 R h c
\end{array}
\)
Thus, transition III gives largest amount of energy.

Hint

\(
\text { (d) } E=E_1 / n^2
\)

Energy used for excitation is 12.75 eV,
i.e. \((-13.6+12.75) \mathrm{eV}=-0.85 \mathrm{eV}\)
The photons of energy 12.75 eV can excite the fourth level of H -atom. As spectral lines emitted is \(=\frac{n(n-1)}{2}=\frac{4(3)}{2}=6\)
Therefore, six lines will be emitted.

Hint

(d) At the distance of closed approach \(r\),
\(
\begin{aligned}
K & =\frac{1}{4 \pi \varepsilon_0} \frac{(2 e)(Z e)}{r} \\
\Rightarrow \quad r & =\frac{2 Z e^2}{4 \pi \varepsilon_0 K}
\end{aligned}
\)
where, \(Z e=\) charge of the nucleus, \(2 e=\) charge of the alpha particle
and \(K=\) kinetic energy of the alpha particle.
\(
\because \quad K=\frac{p^2}{2 m}
\)
where, \(p\) is the momentum of the \(\alpha\)-particle and \(m\) is the mass of the electron.
\(
\begin{aligned}
\therefore \quad r & =\frac{2 Z e^2 2 m}{4 \pi \varepsilon_0 p^2} \text { or } r \propto \frac{1}{p^2} \\
\frac{r^{\prime}}{r} & =\left(\frac{p}{p^{\prime}}\right)^2=\left(\frac{p}{2 p}\right)^2=\frac{1}{4} \Rightarrow r^{\prime}=\frac{r}{4}
\end{aligned}
\)

Hint

(a) According to kinetic interpretation of temperature,
\(
\mathrm{KE}=\left(\frac{1}{2} m v^2\right)=\frac{3}{2} k T
\)
\(
\begin{aligned}
& \Rightarrow \quad 10.2 \times 1.6 \times 10^{-19}=\frac{3}{2} \times\left(1.38 \times 10^{-23}\right) T \\
& \Rightarrow \quad T=7.9 \times 10^4 \mathrm{~K}
\end{aligned}
\)

Hint

\(
\text { (c) } \frac{1}{\lambda}=R Z^2\left(\frac{1}{1^2}-\frac{1}{2^2}\right)
\)
\(
\begin{aligned}
& \frac{1}{\lambda_1}=R(1)^2(3 / 4), \frac{1}{\lambda_2}=R(1)^2(3 / 4) \\
& \frac{1}{\lambda_3}=R(2)^2(3 / 4), \frac{1}{\lambda_4}=R 3^2(3 / 4) \\
& \frac{1}{\lambda_1}=\frac{1}{\lambda_2}=\frac{1}{4 \lambda_3}=\frac{1}{9 \lambda_4} \\
& \lambda_1=\lambda_2=4 \lambda_3=9 \lambda_4
\end{aligned}
\)
So, option (c) is correct.

Hint

(a) According to Bohr theory,
\(
m v r=n \frac{h}{2 \pi} \Rightarrow v=\frac{n h}{2 \pi m r}
\)
and
\(
\frac{m v^2}{r} \propto \frac{k}{r} \Rightarrow \frac{m}{r}\left(\frac{n^2 h^2}{4 \pi^2 m^2 r^2}\right) \propto \frac{k}{r}
\)
\(
\Rightarrow \quad r_n \propto n
\)
Kinetic energy of the electron,
\(
T=\frac{1}{2} m v^2=\frac{1}{2} m\left(\frac{n^2 h^2}{4 \pi^2 m^2 r^2}\right) \Rightarrow T_n \propto \frac{n^2}{r^2}
\)
But as \(r \propto n\), therefore \(T \propto n^0\)

Hint

(a) Total energy given to electron in 4th orbit, \(E=15 \mathrm{eV}\)
Energy of \(n\)th orbit of H -atom,
\(
E^{\prime}=-\frac{13.6}{n^2} \mathrm{eV}
\)
So, energy of 4th orbit of H -atom, \(E^{\prime}=\frac{-13.6}{4^2}=-0.85 \mathrm{eV}\)
\(\therefore\) Ionisation energy of 4th orbit \(=\left|-E^{\prime}\right|\)
\(
=|-0.85 \mathrm{eV}|=0.85 \mathrm{eV}
\)

Hint

(a) In an atom, electron revolves around nucleus. For this, required centripetal forces is provided by electrostatic force of attraction between electron and nucleus. If electrons were stationary, then due to this force, they would fall into the nucleus.
Thus, both Assertion and Reason are correct and Reason is the correct explanation of Assertion.

Hint

(d) To find the number of spectral lines in a hydrogen atom, we can use the formula for the number of spectral lines produced when an electron transitions between energy levels. The formula is:
Number of spectral lines \(=\frac{n(n-1)}{2}\)
where \(n\) is the principal quantum number of the excited state.
Identify the Principal Quantum Number ( \(\mathbf{n}\) ):
The principal quantum number \(n\) can take any positive integer value starting from 1 (i.e., \(n=1,2,3, \ldots\) ).
For the hydrogen atom, there is no upper limit to the value of \(n\) since it can theoretically go to infinity.

Hint

(c) Energy of electron in ground state \((n=1)\) of hydrogen atom,
\(
E_1=-13.6 \mathrm{eV} \quad\left(\because E_n=\frac{-13.6}{n^2} \mathrm{eV}\right)
\)
\(\therefore\) Energy of electron in first excited state \((n=2)\) of hydrogen atom,
\(
E_2=\frac{-13.6}{2^2}=-3.4 \mathrm{eV}
\)
Hence, the excitation energy of hydrogen atom in first excited state,
\(
E=E_2-E_1=-3.4-(-13.6)=10.2 \mathrm{eV}
\)
Therefore, excitation potential is 10.2 V.

Hint

(b) Number of lines emitted \(=\frac{n(n-1)}{2}\) where, \(n=\) number of orbit from which transition takes place.
\(
\Rightarrow \quad 6=\frac{n(n-1)}{2} \text { or } n=4
\)
\(\therefore\) The wavelength of emitted radiations will be maximum for transition \(n=4\) to \(n=3\).

Hint

(d) The energy of the photon,
\(
E=\frac{h c}{\lambda}=\frac{4 \times 10^{-15} \times 3 \times 10^8}{300 \times 10^{-9}}=4 \mathrm{eV}
\)
The ionisation energy is 13.6 eV which is greater than the energy of photon, so electron cannot be knocked out or into excited state. Thus, it will keep orbiting in ground state of the atom.

Hint

(a) Given, \(h c=1240 \mathrm{eV}-\mathrm{nm}\)
For least energetic photon, \(E=E_2-E_1=10.2 \mathrm{eV}\)
We know that, \(E=\frac{h c}{\lambda}\)
\(
\begin{aligned}
\lambda & =\frac{h c}{E} \\
\Rightarrow \quad \lambda & =\frac{1240}{10.2} \\
\lambda & =121.57 \approx 122 \mathrm{~nm}
\end{aligned}
\)

Hint

(c) de-Broglie wavelength, \(\lambda=\frac{h}{p} \dots(i)\)
According to Bohr’s quantisation condition,
\(
m v r_n=\frac{n h}{2 \pi} \Rightarrow p r_n=\frac{n h}{2 \pi} \dots(ii)
\)
From Eqs. (i) and (ii), we get
\(
\lambda=\frac{h \times 2 \pi r_n}{n h} \Rightarrow \lambda=\frac{2 \pi r_n}{n}
\)
For fourth orbit \((n=4) \Rightarrow \lambda=\frac{2 \pi r_4}{4} \dots(iii)\)
Moreover, \(r \propto n^2 \Rightarrow \frac{r_1}{r_4}=\frac{(1)^2}{(4)^2} \Rightarrow r_4=16 r\)
Substituting the value of \(r_4\) in Eq. (iii), we get
\(
\lambda=\frac{2 \pi\left[16 r_1\right]}{4}=8 \pi r_1=8 \pi r \left(\because \text { given, } r_1=r\right)
\)

Hint

\(
\begin{aligned}
&\text { (c) The energy of the photon }=R h c\left(\frac{1}{1^2}-\frac{1}{n^2}\right)\\
&\begin{aligned}
\Rightarrow \quad 13.6\left(\frac{1}{1}-\frac{1}{4}\right) & =13.6\left(\frac{4-1}{4}\right) \\
& =13.6 \times \frac{3}{4}=\frac{40.8}{4}=10.2 \mathrm{eV}
\end{aligned}
\end{aligned}
\)

Hint

(d) When an electron jumps from orbit \(n_i\) to orbit \(n_f\left(n_i>n_f\right)\), the frequency of emitted photon is given by
\(
\Rightarrow \quad \begin{aligned}
& \nu=c R\left(\frac{1}{n_f^2}-\frac{1}{n_i^2}\right) \\
& \nu=c R\left(\frac{1}{2^2}-\frac{1}{3^2}\right) \\
& \nu=c R\left(\frac{5}{36}\right) \Rightarrow \nu=\frac{5 c R}{36}
\end{aligned}
\)

Hint

(c) Rutherford’s model states that an atom is a sphere of diameter about \(10^{-10} \mathrm{~m}\) with whole of its positive charge concentrated at the centre and electron revolves around this centre.

Hint

(c) According to Bohr’s model, electrons can revolve those orbits in which their angular momentum is an integral multiple of \(h / 2 \pi\). i.e.
\(
m v r=\frac{n h}{2 \pi}
\)

Hint

(d) The photoelectric effect is possible when light of sufficient energy incident on metal surface. When electron de-excites from \(n=3\) to \(n=2\), the light produces photoelectric effect. So, for the given options, the photoelectric effect is not possible for that level in which energy of emitted light is less than that for \(n=3\) to \(n=2\). Since, \((h \nu)_{4 \rightarrow 3}<(h \nu)_{3 \rightarrow 2}\)
Hence, photoelectric effect is not possible from \(n=4\) to \(n=3\).

Hint

(b) Ionisation energy in ground state \(=24.6 \mathrm{eV}\)
Energy required to remove 2nd electron from \(\mathrm{He}^{2+}\)
\(
=Z^2(13.6) \mathrm{eV}=(2)^2(13.6)=54.4 \mathrm{eV}
\)
So, total energy required \(=24.6+54.4=79 \mathrm{eV}\)

Hint

(a) As, electron de-excites from \(3 E \rightarrow E\)
\(
\begin{array}{lll}
\therefore & h \nu=3 E-E \\
\text { or } & \frac{h c}{\lambda}=2 E \dots(i)
\end{array}
\)
When electron de-excites from \(\frac{5 E}{3} \rightarrow E\)
\(
\frac{h c}{\lambda^{\prime}}=\frac{5 E}{3}-E=\frac{2 E}{3} \dots(ii)
\)
Substituting the value of \(E[latex] from Eq. (i) in above equation
[latex]
\frac{h c}{\lambda^{\prime}}=\frac{2}{3}\left(\frac{h c}{2 \lambda}\right) \Rightarrow \lambda^{\prime}=3 \lambda
\)

Hint

(d) Lyman series.
Explanation:
Lyman series:
This series of hydrogen spectral lines corresponds to transitions of electrons from higher energy levels to the lowest energy level ( \(n=1\) ), and all wavelengths in this series fall within the ultraviolet region.

Balmer series:
This series corresponds to transitions to the \(\mathrm{n}=2\) level, and most of its wavelengths are in the visible region.

Paschen series:
This series corresponds to transitions to the \(\mathrm{n}=3\) level, and it is located in the infrared region.

Brackett series:
Like the Paschen series, the Brackett series also corresponds to transitions to a higher energy level ( \(n=4\) ) and is situated in the infrared region.

Hint

(d) Minimum energy, \(\begin{aligned} E & =13.6 Z^2 \mathrm{eV} \\ & =13.6(3)^2=122.4 \mathrm{eV}\end{aligned}\)

Hint

(a) The wavelength in Balmer series is given by
\(
\begin{array}{rlrl}
& \frac{1}{\lambda} & =R\left(\frac{1}{2^2}-\frac{1}{n^2}\right), \text { where } n=3,4,5 \\
& \frac{1}{\lambda_{\max }} =R\left(\frac{1}{2^2}-\frac{1}{3^2}\right) \\
\Rightarrow \quad \lambda_{\max } =\frac{36}{5 R}=\frac{36}{5 \times 1.097 \times 10^7}=6563 A \\
\text { and } \quad \frac{1}{\lambda_{\min }} =R\left(\frac{1}{2^2}-\frac{1}{\infty^2}\right) \\
\Rightarrow \quad \lambda_{\max } =\frac{4}{R}=\frac{4}{1.097 \times 10^7}=3646 A
\end{array}
\)
The wavelength \(6563 A\) and \(3646 A\) lie in visible region. Therefore, Balmer series lies in visible region.

Hint

(b) By using, \(\frac{1}{\lambda}=R Z^2\left[\frac{1}{n_1^2}-\frac{1}{n_2^2}\right]\)
Lyman series for hydrogen atom, \(\frac{1}{\lambda_{\min }}=R\left[\frac{1}{1^2}-\frac{1}{\infty}\right]=R\)
\(
\frac{1}{\lambda_{\min }}=R \text { or }\left(\lambda_{\min }\right)_{\mathrm{H}}=\frac{1}{R} \dots(i)
\)
Balmer series for another atom,
\(
\begin{aligned}
\quad \frac{1}{\lambda_{\min }} & =R Z^2\left(\frac{1}{2^2}-\frac{1}{\infty}\right)=\frac{R Z^2}{4} \\
\Rightarrow \quad \lambda_{\min } & =\frac{4}{R Z^2} \dots(ii)
\end{aligned}
\)
From Eqs. (i) and (ii), we get
\(
\frac{1}{R}=\frac{4}{R Z^2} \Rightarrow Z=2
\)

Hint

(c) We have, \(F \propto \frac{v^2}{r}\) Also, \(\quad v \propto \frac{1}{n}\) and \(r \propto n^2\)
\(
\Rightarrow \quad F \propto \frac{1}{n^4}
\)

Hint

\(
\text { (a) } \frac{\lambda_{\text {Lyman }}}{\lambda_{\text {Balmer }}}=\frac{\left(\frac{1}{2^2}-\frac{1}{3^2}\right)}{\left(\frac{1}{1^2}-\frac{1}{2^2}\right)}=\frac{5}{27} \quad\left[\because \frac{1}{\lambda}=R\left(\frac{1}{n_f^2}-\frac{1}{n_i^2}\right)\right]
\)
\(
\lambda_{\text {Lyman }}=\frac{5}{27} \times \lambda_{\text {Balmer }}=\frac{5}{27} \times 6563=1215.4 Å
\)

Hint

\(
\begin{aligned}
&\text { (c) Time period of electron, }\\
&T=\frac{2 \pi r}{v}=\frac{\frac{2 \pi \times \varepsilon_0 n^2 h^2}{\pi m e^2}}{\frac{e^2}{2 \varepsilon_0 n h}}
\end{aligned}
\)
\(
\begin{array}{ll}
\text { Current, } & T=\frac{4 \varepsilon_0^2 n^3 h^3}{m e^4} \\
\Rightarrow & i=\frac{e}{T}=\frac{e}{\frac{4 \varepsilon_0^2 n^3 h^3}{m e^4}} \\
\text { i.e. } & i \propto e^5
\end{array}
\)

Hint

(a) Maximum energy is liberated for transition \(E_n \rightarrow E_1\) and minimum energy for \(E_n \rightarrow E_{n-1}\).
\(
\begin{aligned}
& \text { Hence, } \frac{E_1}{n^2}-E_1=52.224 \mathrm{eV} \dots(i) \\
& \text { and } \frac{E_1}{n^2}-\frac{E_1}{(n-1)^2}=1.224 \mathrm{eV} \dots(ii)
\end{aligned}
\)
Solving Eqs. (i) and (ii), we get
\(
E_1=-54.4 \mathrm{eV} \text { and } n=5
\)
\(
\begin{array}{lrl}
\text { But } & E_1 & =-\frac{13.6 Z^2}{1^2} \\
\therefore & -54.4 & =-\frac{13.6}{1^2} Z^2 \\
\Rightarrow & Z & =2
\end{array}
\)