Past NEET Entrance Papers

Hint

(a)

\(
\begin{aligned}
& \text { Energy difference } \Delta E=\frac{h c}{\lambda} \\
& \therefore \lambda \propto \frac{1}{\Delta E} \\
& (\Delta E)_{6-2}>(\Delta E)_{5-2}>(\Delta E)_{4-2}>(\Delta E)_{3-2} \\
& \lambda_{6-2}<\lambda_{5-2}<\lambda_{4-2}<\lambda_{3-2}
\end{aligned}
\)

Hint

(c) Statement I: “Atoms are electrically neutral as they contain equal number of positive and negative charges.”

This statement is correct. Atoms are composed of protons, neutrons, and electrons. Protons have a positive charge, electrons have a negative charge, and neutrons have no charge. In a neutral atom, the number of protons (positive charges) equals the number of electrons (negative charges), thus making the atom electrically neutral. The positive and negative charges balance each other, resulting in no overall charge.

Statement II: “Atoms of each element are stable and emit their characteristic spectrum.”

This statement requires modification for accuracy. Atoms emit their characteristic spectrum when they are excited and then return to a lower energy state. The emission of a spectrum is not linked directly to the stability of an atom; it is linked to the electronic transitions within the atom. The term “stable” in relation to atoms typically refers to the overall energy state of an atom being at its lowest or in a ground state. However, atoms can emit radiation whether they are in a stable ground state or in an excited state. Moreover, not all atoms are inherently stable; some may be radioactive and unstable. The part that atoms emit their characteristic spectrum under certain conditions (like excitation) is correct, but linking this property strictly to stability is misleading.

Based on the analysis:

Statement I is correct.
Statement II is partially correct, as it is accurate that each element can emit a characteristic spectrum when excited, but incorrect in generalizing that all such atoms are stable.

Hint

(c)

Lyman series (B):
corresponds to transitions where the final energy level \(\left(n^{\prime}\right)\) is 1 , so the wave number expression is \(R\left(1 / 1^2-1 / n^2\right)\), which is option I.

Balmer series (A):
corresponds to transitions where the final energy level \(\left(n^{\prime}\right)\) is 2 , so the wave number expression is \(R\left(1 / 2^2-1 / n^2\right)\), which is option IV.

Brackett series (C):
corresponds to transitions where the final energy level \(\left(n^{\prime}\right)\) is 4 , so the wave number expression is \(R\left(1 / 4^2-1 / n^2\right)\), which is option II.

Pfund series (D):
corresponds to transitions where the final energy level \(\left(n^{\prime}\right)\) is 5 , so the wave number expression is \(\mathrm{R}\left(1 / 5^2-1 / \mathrm{n}^2\right)\), which is option III.

Hint

(b) The spectral series corresponds to groups of wavelengths that are produced when electrons in an atom make transitions between energy levels. Each series is named after the scientist who discovered it and is characterized by the energy level to which the electrons transition.

The transitions to the level \(n_1=4\) define the Brackett series. Electrons that fall to the fourth energy level (from higher levels with \(n_2=5,6, \ldots\) ) emit or absorb radiation in the infrared region of the electromagnetic spectrum.

Here’s a quick summary of the different spectral series:
Pfund series: Transitions to \(n_1=5\)
Brackett series: Transitions to \(n_1=4\)
Lyman series: Transitions to \(n_1=1\)
Balmer series: Transitions to \(n_1=2\)

Hint

(d)

\(
\begin{aligned}
& h \frac{c}{\lambda_1}=\frac{-5 E}{2}+4 E=\frac{3}{2} E \dots(1) \\
& h \frac{c}{\lambda_2}=-2 E+3 E=E \dots(2) \\
& h \frac{c}{\lambda_3}=-2 E+4 E=2 E \dots(3)
\end{aligned}
\)
Comparing (2) and (3)
\(
\frac{1}{\lambda_3}=\frac{2}{\lambda_2} \quad \lambda_2=2 \lambda_3
\)
Comparing (1) and (2)
\(
3 \lambda_1=2 \lambda_2 \quad \lambda_1<\lambda_2
\)

Hint

(c) The wavelength of the emitted light during a transition in the hydrogen atom can be calculated using the Rydberg formula:
\(
\frac{1}{\lambda}=R\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)
\)
where:
\(R\) is the Rydberg constant,
\(n_1\) is the lower energy level,
\(n_2\) is the higher energy level.

Determine the Values for the Brackett Series:
For the Brackett series, the lowest energy level is \(n_1=4\) and the transitions can go to higher levels starting from \(n_2=5,6, \ldots\) up to infinity. The shortest wavelength corresponds to the transition from \(n_2=\infty\) to \(n_1=4\).

Apply the Rydberg Formula:
Substituting the values into the Rydberg formula for the Brackett series:
\(
\frac{1}{\lambda_{\text {Brackett }}}=R\left(\frac{1}{4^2}-\frac{1}{\infty^2}\right)
\)
Since \(\frac{1}{\infty^2}=0\), the equation simplifies to:
\(
\frac{1}{\lambda_{\text {Brackett }}}=R\left(\frac{1}{16}\right)
\)

Rearranging the equation gives:
\(
\lambda_{\text {Brackett }}=\frac{16}{R}
\)

Relate to the Wavelength in the Balmer Series:
From the Balmer series, we know that the shortest wavelength \(\lambda_{\text {Balmer }}\) corresponds to the transition from \(n_2=\infty\) to \(n_1=2\) :
\(
\frac{1}{\lambda_{\text {Balmer }}}=R\left(\frac{1}{2^2}-0\right)=\frac{R}{4}
\)
Thus,
\(
\lambda_{\text {Balmer }}=\frac{4}{R}
\)

Relate the Two Wavelengths:
Since we have:
\(
\lambda_{\text {Brackett }}=\frac{16}{R}=4 \cdot \frac{4}{R}=4 \cdot \lambda_{\text {Balmer }}
\)

Hint

(a) The radius of the \(n[latex]-th orbit is given by [latex]r_n=n^2 r_1\).
Substitute \(n=3\) and \(r_1=5.3 \times 10^{-11} \mathrm{~m}\) into the formula \(r_n=n^2 r_1\).
\(r_3=3^2 \times 5.3 \times 10^{-11} \mathrm{~m}\)
\(r_3=9 \times 5.3 \times 10^{-11} \mathrm{~m}\)
\(r_3=47.7 \times 10^{-11} \mathrm{~m}\)
Use the conversion factor \(1 Å ̈=10^{-10} \mathrm{~m}\).
\(r_3=47.7 \times 10^{-11} \mathrm{~m} \times \frac{1 \AA}{10^{-10} \mathrm{~m}}\)
\(r_3=4.77 \times 10^{-10} \mathrm{~m} \times \frac{1}{10^{-10}} \frac{Å}{\mathrm{~m}}\)
\(r_3=4.77 Å\)
The radius of the third allowed orbit of a hydrogen atom is \(4.77 Å\).

Hint

(c) The energy levels of a hydrogen atom are given by the formula :
\(
E_n=-\frac{13.6 \mathrm{eV}}{n^2}
\)
where \(E_n\) is the energy of the \(n\)-th energy level.
The ground state of hydrogen \((n=1)\) has an energy of -13.6 eV as mentioned, which means that it would take +13.6 eV to ionize it (remove the electron completely) from this state, since ionization implies moving the electron to a state of zero energy.
The second excited state of hydrogen is when \(n=3[latex] (as [latex]n=1\) is the ground state and \(n=2\) is the first excited state). Thus, the energy of the second excited state is :
\(
E_3=-\frac{13.6 \mathrm{eV}}{3^2}=-\frac{13.6 \mathrm{eV}}{9}=-1.51 \mathrm{eV}
\)
Since ionization implies moving the electron from its current energy level to 0 energy, the energy required to ionize the atom from this state is the absolute value of its current energy state. So, it will take +1.51 eV to ionize a hydrogen atom from its second excited state.

Hint

(c) The Lyman series corresponds to transitions to the \(n=1\) energy level.

The Rydberg formula is given by \(\frac{1}{\lambda}=R_H\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)\), where \(R_H \approx 1.097 \times 10^7 \mathrm{~m}^{-1}\) is the Rydberg constant, \(n_1\) and \(n_2\) are the principal quantum numbers of the initial and final energy levels, and \(\lambda\) is the wavelength of the emitted photon.
The Lyman series corresponds to \(n_1=1\) and \(n_2>1\).
The ultraviolet region of the electromagnetic spectrum corresponds to wavelengths between approximately 10 nm and 400 nm .

How to solve?
Calculate the shortest and longest wavelengths in the Lyman series using the Rydberg formula and determine the region of the electromagnetic spectrum they belong to.

Step 1: Calculate the shortest wavelength in the Lyman series
The shortest wavelength corresponds to the largest energy transition, which occurs when \(n_2=\infty\).
Use the Rydberg formula:
\(\frac{1}{\lambda}=R_H\left(\frac{1}{1^2}-\frac{1}{\infty^2}\right)\)
\(\frac{1}{\lambda}=R_H\)
\(\lambda=\frac{1}{R_H}\)
\(\lambda=\frac{1}{1.097 \times 10^7 m^{-1}}\)
\(\lambda \approx 91.1 \times 10^{-9} \mathrm{~m}=91.1 \mathrm{~nm}\)
Step 2: Calculate the longest wavelength in the Lyman series
The longest wavelength corresponds to the smallest energy transition, which occurs when \(n_2=2\).

Use the Rydberg formula:
\(\frac{1}{\lambda}=R_H\left(\frac{1}{1^2}-\frac{1}{2^2}\right)\)
\(\frac{1}{\lambda}=R_H\left(1-\frac{1}{4}\right)\)
\(\frac{1}{\lambda}=R_H\left(\frac{3}{4}\right)\)
\(\lambda=\frac{4}{3 R_H}\)
\(\lambda=\frac{4}{3 \times 1.097 \times 10^7 m^{-1}}\)
\(\lambda \approx 121.5 \times 10^{-9} \mathrm{~m}=121.5 \mathrm{~nm}\)

Step 3: Determine the region of the electromagnetic spectrum
The calculated wavelengths, 91.1 nm and 121.5 nm , fall within the ultraviolet region ( 10 nm to 400 nm ).

Solution: The Lyman series of the hydrogen atom appears in the ultraviolet region.

Hint

(a) Angular momentum \(L=1.5 \frac{h}{\pi}\)
Angular momentum is quantized: \(L=n \frac{h}{2 \pi}\)
Energy levels of hydrogen atom: \(E_n=-\frac{13.6}{n^2} \mathrm{eV}\)
How to solve?
Find the principal quantum number \(n\) from the angular momentum, then calculate the energy using the energy level formula.

Step 1:
Find the principal quantum number \(n\)
Equate the given angular momentum to the quantized angular momentum formula:
\(1.5 \frac{h}{\pi}=n \frac{h}{2 \pi}\)

Solve for \(n\) :
\(n=\frac{1.5 \frac{h}{\pi}}{\frac{h}{2 \pi}}\)
\(n=1.5 \cdot 2\)
\(n=3\)

Step 2:
Use the energy level formula with \(n=3\) :
\(E_3=-\frac{13.6}{3^2} \mathrm{eV}\)
\(E_3=-\frac{13.6}{9} \mathrm{eV}\)
\(E_3 \approx-1.5 \mathrm{eV}\)
Solution: The energy of the electron in the given orbit is approximately -1.5 eV.

Hint

(a) \(T_1\) is the energy of an electron in the first excited state. \({T}_2\) is the energy of an electron in the second excited state.
The energy of an electron in the \(n\)-th orbit of a hydrogen atom is given by \(T_n=-\frac{13.6}{n^2} \mathrm{eV}\).
The first excited state corresponds to \(n=2\).
The second excited state corresponds to \(n=3\).
How to solve?
Calculate \(T_1\) and \(T_2\) using the formula for the energy of an electron in the \(n\)-th orbit, then find the ratio \(T_1: T_2\).

\(T_1\) = energy of electron in first excited state
\(T_2\) = energy of electron in second excited state
From equation (1) we can say:
\(T_1=-13.6 / n^2\), where \(n=2\) for first excited state
\(
T_1=-13.6 / 2^2=-13.6 / 4 \dots(1)
\)
\(
T_2({n}=3)=-13.6 / 3^2=-13.6 / 9 \dots(2)
\)
From equation (1) and (2) then we get:
\(
T_1 / T_2=(1 / 4) /(1 / 9)=9 / 4
\)
\(T_1: T_2 = 9:4\)

Hint

(d) The electron is in the first excited state, \(n_1=2\).
The electron is in the second excited state, \(n_2=3\).
According to Bohr’s model, the angular momentum of an electron is quantized: \(L=n \frac{h}{2 \pi}\), where \(n\) is the principal quantum number.
Step 1:
Calculate the angular momentum \(L_1\) for the first excited state ( \(n_1=2\) ).
\(
\begin{aligned}
& L_1=n_1 \frac{h}{2 \pi} \\
& L_1=2 \frac{h}{2 \pi} \\
& L_1=\frac{h}{\pi}
\end{aligned}
\)

Step 2:
\(
\begin{aligned}
& L_2=n_2 \frac{h}{2 \pi} \\
& L_2=3 \frac{h}{2 \pi} \\
& L_2=\frac{3 h}{2 \pi}
\end{aligned}
\)

Step 3: Find the ratio \(L_1: L_2\).

Solution: The ratio of the orbital angular momentum \(L_1: L_2\) is \(2: 3\).

Hint

(a) \(R_1\) is the radius of the second orbit, so \(n=2\).
\(R_2\) is the radius of the fourth orbit, so \(n=4\).
The radius of the \(n^{\text {th }}\) orbit in Bohr’s model is given by \(R_n=n^2 a_0\), where \(a_0\) is the Bohr radius.
Radius of Bohr’s orbit depends on principal quantum number \((\mathrm{n})\) as
\(
R \propto n^2
\)
Now, \(\frac{R_1}{R_2}=\frac{(2)^2}{(4)^2}=\frac{1}{4}=0.25\)

Hint

(c) Explanation: The Bohr model is only valid for single-electron systems, meaning it only accurately describes the behavior of an atom with only one electron. A singly ionized neon atom ( \(\mathrm{Ne}^{+}\)) has more than one electron, so the Bohr model cannot accurately predict its behavior.

Why other options are incorrect:
(A) Singly ionised helium atom \(\left(\mathrm{He}^{+}\right)\): This is a single-electron system, so the Bohr model is valid for it.
(B) Deuteron atom: A deuteron is an isotope of hydrogen with one proton and one neutron. Since it has only one electron, the Bohr model is applicable.
(D) Hydrogen atom: The Bohr model was originally developed to explain the behavior of the hydrogen atom, which is a single-electron system.

Hint

(b) The formula for the total energy of an electron in the \(n^{\text {th }}\) orbit of a hydrogen atom is:
\(E_n=-\frac{13.6}{n^2} \mathrm{eV}\)
where \(n\) is the principal quantum number.

Hint

(a) \(E=m c^2=0.5 \times 10^{-3} \times\left(3 \times 10^8\right)^2=4.5 \times 10^{13} J\)

Hint

(d) Total energy \(E=-3.4 \mathrm{eV}\)
Kinetic energy \(K\) is the negative of the total energy \(E: K=-E\).
Potential energy \(U\) is twice the total energy \(E: U=2 E\).
Use the formula \(K=-E\).
Substitute the given value of \(E\) :
\(K=-(-3.4 \mathrm{eV})\)
\(K=3.4 \mathrm{eV}\)

Use the formula \(U=2 E\).
Substitute the given value of \(E\) :
\(U=2(-3.4 \mathrm{eV})\)
\(U=-6.8 \mathrm{eV}\)

Solution: The kinetic energy is 3.4 eV and the potential energy is -6.8 eV.

Hint

(b) \(\alpha \text { – particle is nucleus of Helium which has two protons and two neutrons. }\)

Hint

(d) Concept:
The Bohr radius is written as
\(
r=\frac{n^2 h^2}{4 \pi^2 m k z e^2}
\)
Here we have \(m\) as the mass, \({h}\) as Planck’s constant, \({m}\) is the mass, \({z}\) is the atomic number, and \({e}\) is the electron mass.
The Bohr radius is inversely proportional to the mass of the orbiting particle: \(r \propto \frac{1}{m}\). The ground state energy is directly proportional to the mass of the orbiting particle: \(E \propto m\).

Explanation:
The radius of the first Bohr orbit for an electron in a hydrogen atom: \(r_e=0.51 Å\)
The ground state energy of an electron in a hydrogen atom: \(E_e=-13.6 {eV}\)
The mass of the muon: \(m_\mu=207 m_e\)
The charge of the muon is the same as the electron.
The new Bohr radius \(r_\mu\) is related to the electron Bohr radius \(r_e\) by the inverse ratio of their masses:
\(r_\mu=r_e \frac{m_e}{m_\mu}\)
\(r_\mu=r_e \frac{m_e}{207 m_e}\)
\(r_\mu=\frac{r_e}{207}\)
\(r_\mu=\frac{0.51 Å}{207}=2.56 \times 10^{-13} \mathrm{~m}\)

The new ground state energy \(E_\mu\) is related to the electron ground state energy \(E_e\) by the ratio of their masses:
\(E_\mu=E_e \frac{m_\mu}{m_e}\)
\(E_\mu=E_e \frac{207 m_e}{m_e}\)
\(E_\mu=207 E_e\)
\(E_\mu=207 \times(-13.6 \mathrm{eV})\)
\({E}_\mu=-2815.2 \mathrm{eV}\)
\(E_\mu=-2.8152 \mathrm{keV}\)
\(E_\mu \approx-2.8 \mathrm{keV}\)

Hint

(b) The total energy \(E\) is given by \(E=K E+P E\).
Since \(K E=-\frac{1}{2} P E\), then \(P E=-2 K E\).
Substituting \(P E\) into the total energy equation: \(E=K E-2 K E=-K E\).
The ratio of kinetic energy to total energy is \(\frac{K E}{E}\). Since \(E=-K E\), the ratio is \(\frac{K E}{-K E}=\frac{1}{-1}\).
The ratio of kinetic energy to the total energy of an electron in a Bohr orbit of the hydrogen atom is \(1:-1\).

Hint

(b) For last line of Blamer series, \(n_1=2, n_2=\infty\)
\(
\frac{1}{\lambda_B}=R\left(\frac{1}{2^2}-\frac{1}{\infty}\right)=\frac{R}{4} \ldots(i)
\)
for last line of Lyman series, \(n_1=1, n_2=\infty\)
\(
\frac{1}{\lambda_L}=R\left(\frac{1}{1^2}-\frac{1}{\infty}\right)=R \ldots \ldots \ldots .(i i)
\)
Dividing (ii) by (i), we get
\(
\frac{\lambda_B}{\lambda_L}=4
\)

Hint

(d) When electron jumps from higher orbit to lower orbit then, wavelength of emitted photon is given by,
\(
\frac{1}{\lambda}=R\left(\frac{1}{n_f^2}-\frac{1}{n_i^2}\right)
\)
On jumping from \(3^{\text {rd }}\) orbit to \(2^{\text {nd }}\) orbit,
\(
\frac{1}{\lambda}=R\left(\frac{1}{2^2}-\frac{1}{3^2}\right)=\frac{5 R}{36}
\)
On jumping from \(4^{\text {th }}\) orbit to \(3^{\text {rd }}\) orbit,
\(
\begin{aligned}
& \frac{1}{\lambda^{\prime}}=R\left(\frac{1}{3^2}-\frac{1}{4^2}\right)=\frac{7 R}{144} \\
& \therefore \lambda^{\prime}=\frac{144}{7} \times \frac{5 \lambda}{36}=\frac{20 \lambda}{7}
\end{aligned}
\)

Hint

(b) The wave number for the Balmer series is given by:
\(\frac{1}{\lambda}=R\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)\)
where \(n_1=2\) for the Balmer series.

For the last line of the Balmer series, \(n_2=\infty\).
Substitute \(n_1=2\) and \(n_2=\infty\) into the formula:
\(\frac{1}{\lambda}=R\left(\frac{1}{2^2}-\frac{1}{\infty^2}\right)\)
\(\frac{1}{\lambda}=R\left(\frac{1}{4}-0\right)\)
\(\frac{1}{\lambda}=\frac{R}{4}\)
Given \(R=10^7 \mathrm{~m}^{-1}\), substitute it into the equation:
\(\frac{1}{\lambda}=\frac{10^7 \mathrm{~m}^{-1}}{4}\)
\(\frac{1}{\lambda}=0.25 \times 10^7 \mathrm{~m}^{-1}\)

Hint

(d) Mass of \(\alpha\)-particle: \(m\)
Velocity of \(\alpha\)-particle: \(v\)
Charge of heavy nucleus: \(Z e\)
Charge of \(\alpha\)-particle: \(2 e\)
The kinetic energy of the \(\alpha\)-particle is converted into potential energy at the distance of closest approach.
Kinetic energy is given by \(K E=\frac{1}{2} m v^2\).
Electrostatic potential energy is given by \(P E=\frac{k q_1 q_2}{r}\), where \(k=\frac{1}{4 \pi \epsilon_0}\).
Step 1:
Initial kinetic energy of the \(\alpha\)-particle:
\(K E=\frac{1}{2} m v^2\)
Potential energy at the distance of closest approach \(r\) :
\(P E=\frac{k(2 e)(Z e)}{r}=\frac{2 k Z e^2}{r}\)
At the distance of closest approach, \(K E=P E\) :
\(\frac{1}{2} m v^2=\frac{2 k Z e^2}{r}\)

Step 2:
Rearrange the equation to solve for \(r\) :
\(r=\frac{4 k Z e^2}{m v^2}\)

Step 3:
From the equation \(r=\frac{4 k Z e^2}{m v^2}\), it is clear that \(r\) is inversely proportional to \(m\) :
\(r \propto \frac{1}{m}\)

Solution: The distance of closest approach depends on \(m\) as \(\frac{1}{m}\).

Alternate: At the closest distance of approach, the kinetic energy of the particle will convert completely into electrostatic potential energy.
\(
\text { Kinetic energy }=\frac{1}{2} m v^2
\)
\(
\text { Potential energy }=\frac{K Q q}{r}
\)
\(
\begin{aligned}
& \therefore \frac{1}{2} m v^2=\frac{K Q q}{r} \\
& \Rightarrow r \propto \frac{1}{m}
\end{aligned}
\)

Hint

(b) Rydberg formula: \(\frac{1}{\lambda}=R\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)\), where \(R\) is the Rydberg constant, \(\lambda\) is the wavelength, \(n_1\) and \(n_2\) are integers with \(n_2>n_1\).
For the Lyman series, \(n_1=1\).
For the Balmer series, \(n_1=2\).
The longest wavelength corresponds to the smallest energy difference, thus the smallest difference between \(n_2\) and \(n_1\).
Apply the Rydberg formula to find the longest wavelengths for the Lyman and Balmer series, then calculate their ratio.

Step 1: Find the longest wavelength for the Lyman series
For the longest wavelength in the Lyman series, \(n_1=1\) and \(n_2=2\).
Apply the Rydberg formula:
\(\frac{1}{\lambda_L}=R\left(\frac{1}{1^2}-\frac{1}{2^2}\right)\)
\(\frac{1}{\lambda_L}=R\left(1-\frac{1}{4}\right)\)
\(\frac{1}{\lambda_L}=R\left(\frac{3}{4}\right)\)
\(\lambda_L=\frac{4}{3 R}\)

Step 2: For the longest wavelength in the Balmer series, \(n_1=2\) and \(n_2=3\).
Apply the Rydberg formula:
\(\frac{1}{\lambda_B}=R\left(\frac{1}{2^2}-\frac{1}{3^2}\right)\)
\(\frac{1}{\lambda_B}=R\left(\frac{1}{4}-\frac{1}{9}\right)\)
\(\frac{1}{\lambda_B}=R\left(\frac{9-4}{36}\right)\)
\(\frac{1}{\lambda_B}=R\left(\frac{5}{36}\right)\)
\(\lambda_B=\frac{36}{5 R}\)

Step 3: Divide the longest wavelength in the Lyman series by the longest wavelength in the Balmer series:
\(\frac{\lambda_L}{\lambda_B}=\frac{\frac{4}{3 R}}{\frac{36}{5 R}}=\frac{5}{27}\)

Hint

(b) Orbit number: \(n=3\)
Atomic number of Helium: \(\boldsymbol{Z}=2\)
Planck’s constant: \(h=6.6 \times 10^{-34} \mathrm{~J} \mathrm{~s}\)

Bohr’s quantization condition: \(m v r=n \frac{h}{2 \pi}\)
Speed of electron in nth orbit: \(v=\frac{2 \pi Z e^2}{n h 4 \pi \epsilon_0}=\frac{Z}{n} \times 2.18 \times 10^6 \mathrm{~m} / \mathrm{s}\)

Use the formula \(v=\frac{Z}{n} \times 2.18 \times 10^6 \mathrm{~m} / \mathrm{s}\)
Substitute \(Z=2[latex] and [latex]n=3\)
\(v=\frac{2}{3} \times 2.18 \times 10^6 \mathrm{~m} / \mathrm{s}\)
\(v=1.453 \times 10^6 \mathrm{~m} / \mathrm{s}\)
\(v \approx 1.46 \times 10^6 \mathrm{~m} / \mathrm{s}\)

Solution: The speed of the electron in the third orbit of \(\mathrm{He}^{+}\)is approximately \(1.46 \times 10^6 \mathrm{~m} / \mathrm{s}\).

Hint

(b) The initial kinetic energy is the sum of the kinetic energies of both particles:
\(K E_{\text {initial }}=\frac{1}{2} m_1 u_1^2+\frac{1}{2} m_2 u_2^2\)
The final kinetic energy is the sum of the kinetic energies of both particles after the collision:
\(K E_{\text {final }}=\frac{1}{2} m_1 v_1^2+\frac{1}{2} m_2 v_2^2\)
The initial kinetic energy minus the energy absorbed equals the final kinetic energy:
\(K E_{\text {initial }}-\varepsilon=K E_{\text {final }}\)
\(\frac{1}{2} m_1 u_1^2+\frac{1}{2} m_2 u_2^2-\varepsilon=\frac{1}{2} m_1 v_1^2+\frac{1}{2} m_2 v_2^2\)
The relationship between the initial and final kinetic energies, considering the energy absorbed, is \(\frac{1}{2} m_1 u_1^2+\frac{1}{2} m_2 u_2^2-\varepsilon=\frac{1}{2} m_1 v_1^2+\frac{1}{2} m_2 v_2^2\).

Hint

(c) Wavelength of radiation: \(\lambda=975 Å=975 \times 10^{-10} \mathrm{~m}\)
Hydrogen atom is initially in the ground state \(\left(n_1=1\right)\)

Rydberg formula: \(\frac{1}{\lambda}=R_H\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)\), where \(R_H \approx 1.097 \times 10^7 m^{-1}\) is the Rydberg constant.
Number of spectral lines emitted when an electron transitions from level \(n\) to lower levels is given by: \(N=\frac{n(n-1)}{2}\).
Calculate the final energy level \(n_2\)
Use the Rydberg formula: \(\frac{1}{\lambda}=R_H\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)\).
Substitute the given values: \(\frac{1}{975 \times 10^{-10} \mathrm{~m}}=1.097 \times 10^7 \mathrm{~m}^{-1}\left(\frac{1}{1^2}-\frac{1}{n_2^2}\right)\).
Simplify: \(\frac{1}{975 \times 10^{-10} \times 1.097 \times 10^7}=1-\frac{1}{n_2^2}\).
Calculate: \(\frac{1}{0.107} \approx 1=1-\frac{1}{n_2^2}\).
Rearrange: \(\frac{1}{n_2^2}=1-0.929 \approx 0.071\).
Solve for \(n_2^2: n_2^2 \approx \frac{1}{0.071} \approx 14.08\).
Take the square root: \(n_2 \approx \sqrt{14.08} \approx 3.75 \approx 4\). Since \(n_2\) must be an integer, \(n_2=4\).
Calculate the number of spectral lines
Use the formula: \(N=\frac{n_2\left(n_2-1\right)}{2}\).
Substitute \(n_2=4: N=\frac{4(4-1)}{2}\).
Calculate: \(N=\frac{4 \times 3}{2}=6\).

Solution: The number of spectral lines emitted is 6.

Hint

(d) According to Bohr’s model, the time period of an electron is proportional to the cube of the principal quantum number: \(T \propto n^3\).
The time period of the electron in the initial state is eight times that in the final state: \(T_1=8 T_2\).
Since \(T \propto n^3\), we can write \(T_1=k n_1^3\) and \(T_2=k n_2^3\), where \(k\) is a constant of proportionality.
Given \(T_1=8 T_2\), substitute the expressions for \(T_1\) and \(T_2\) :
\(k n_1^3=8 k n_2^3\)
Divide both sides by \(k\) :
\(n_1^3=8 n_2^3\)
Take the cube root of both sides of the equation:
\(\sqrt{n_1^3}=\sqrt{8 n_2^3}\)
\(n_1=2 n_2\)
Since \(n_1\) and \(n_2\) are integers, we can find possible pairs that satisfy \(n_1=2 n_2\).
If \(n_2=1\), then \(n_1=2(1)=2\).
If \(n_2=2\), then \(n_1=2(2)=4\).
If \(n_2=3\), then \(n_1=2(3)=6\), and so on.
The most common and simplest solution is \(n_1=4\) and \(n_2=2\).

Hint

(b) The wavelength of emission is given by
\(
\Rightarrow \frac{1}{\lambda}=R\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)
\)
Case 1:
Third excited state means \({n}_2=4\)
Second excited state means \(n_1=3\)
\(
\begin{aligned}
& \Rightarrow \frac{1}{\lambda_1}=R\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right) \\
& \Rightarrow \frac{1}{\lambda_1}=R\left(\frac{1}{3^2}-\frac{1}{4^2}\right)=R\left(\frac{1}{9}-\frac{1}{16}\right)=R\left(\frac{16-9}{16 \times 9}\right)=\frac{7 R}{144} \\
& \Rightarrow \lambda_1=\frac{144}{7 R} \dots(1)
\end{aligned}
\)

Case 2:
In the case of \(\lambda_2 \quad n_1=2\)
Value of \(n_2=3\)
\(
\begin{aligned}
& \Rightarrow \frac{1}{\lambda_2}=R\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right) \\
& \Rightarrow \frac{1}{\lambda_2}=R\left(\frac{1}{2^2}-\frac{1}{3^2}\right)=R\left(\frac{1}{4}-\frac{1}{9}\right)=R\left(\frac{9-4}{36}\right)=\frac{5 R}{36} \\
& \Rightarrow \lambda_2=\frac{36}{5 R} \dots(2)
\end{aligned}
\)
Divide equation 1 and 2 , we get
\(
\Rightarrow \frac{\lambda_1}{\lambda_2}=\frac{\frac{144}{7 R}}{\frac{36}{5 R}}=\frac{20}{7}
\)

Hint

(d) The energy of a photon emitted during an electron transition in a hydrogen atom is given by \(\Delta E=h f=h c / \lambda=\operatorname{Rhc}\left(\frac{1}{n_f^2}-\frac{1}{n_i^2}\right)\).
The momentum of a photon is given by \(p=\frac{E}{c}=\frac{h}{\lambda}\).
Conservation of momentum: \(p_{\text {atom }}=p_{\text {photon }}\).
The energy of the emitted photon is given by:
\(\Delta E=R h c\left(\frac{1}{n_f^2}-\frac{1}{n_i^2}\right)\)
\(\Delta E=\operatorname{Rhc}\left(\frac{1}{1^2}-\frac{1}{5^2}\right)\)
\(\Delta E=R h c\left(1-\frac{1}{25}\right)\)
\(\Delta E=R h c\left(\frac{24}{25}\right)\)
The momentum of the photon is given by:
\(p_{\text {photon }}=\frac{\Delta E}{c}\)
\(p_{\text {photon }}=\frac{R h c}{c} \cdot \frac{24}{25}\)
\(p_{\text {photon }}=R h \cdot \frac{24}{25}\)
By conservation of momentum, the momentum of the atom is equal to the momentum of the photon:
\(p_{\text {atom }}=p_{\text {photon }}\)
\(m v=R h \cdot \frac{24}{25}\)
\(
v=\frac{24 R h}{25 m}
\)

Hint

(c) \(E_n=-\frac{13.6}{n^2} \mathrm{eV}\)
For the ground state ( \(n=1\) ):
\(
E_1=-\frac{13.6}{1^2}=-13.6 \mathrm{eV}
\)
For the first excited state ( \(n=2\) ):
\(
E_2=-\frac{13.6}{2^2}=-\frac{13.6}{4}=-3.4 \mathrm{eV}
\)
The energy released when the electron transitions from the first excited state to the ground state is given by:
\(
\begin{gathered}
E=E_2-E_1=(-3.4)-(-13.6)=-3.4+13.6 \\
=10.2 \mathrm{eV}
\end{gathered}
\)
According to the photoelectric effect, the energy of the incident photon is related to the maximum kinetic energy of the emitted electrons and the work function \((\phi)\) :
\(
E=K E_{\max }+\phi
\)
Rearranging gives:
\(
\phi=E-K E_{\max }
\)
Substituting the values we have:
\(
\phi=10.2 \mathrm{eV}-3.57 \mathrm{eV}=6.63 \mathrm{eV}
\)
The threshold frequency \(\left(\nu_0\right)\) can be calculated using the work function and Planck’s constant ( \(h\) ):
\(
\phi=h \nu_0
\)
Rearranging gives:
\(
\nu_0=\frac{\phi}{h}
\)
Using \(h=6.63 \times 10^{-34} \mathrm{Js}\) and converting \(\phi\) to joules:
\(
\phi=6.63 \mathrm{eV} \times 1.6 \times 10^{-19} \mathrm{~J} / \mathrm{eV}=1.0608 \times 10^{-18} \mathrm{~J}
\)
Now substituting:
\(
\nu_0=\frac{1.0608 \times 10^{-18}}{6.63 \times 10^{-34}} \approx 1.6 \times 10^{15} \mathrm{~Hz}
\)

Hint

(d) Transition \(2 \rightarrow 1\) belongs to the Lyman series (Ultraviolet)
Transition \(3 \rightarrow 2\) and \(5 \rightarrow 2\) belongs to the Balmer series (visible)
Transition 4 \(\rightarrow 3\) belongs to the Paschen series (Infrared)
\(
\begin{array}{|c|c|c|c|}
\hline \text { Series } & {n}_{{f}} & {n}_i & \text { Region } \\
\hline \text { Lyman } & 1 & 2,3,4, \ldots & \text { Ultraviolet } \\
\hline \text { Balmer } & 2 & 3,4,5, \ldots & \text { Visible } \\
\hline \text { Paschen } & 3 & 4,5,6, \ldots & \text { Infrared } \\
\hline \text { Brackett } & 4 & 5.6 .7 \ldots & \text { Infra Red } \\
\hline \text { Pfund } & 5 & 6,7,8, \ldots & \text { Infra Red } \\
\hline
\end{array}
\)

Hint

(c) The Rydberg formula for the wavelength of the emitted light is given by:
\(
\frac{1}{\lambda}=R Z^2\left(\frac{1}{n_f^2}-\frac{1}{n_i^2}\right)
\)
where \(R\) is the Rydberg constant, \(Z\) is the atomic number, \(n_f\) is the final energy level, and \(n_i\) is the initial energy level.
Calculate Wavelength for Lyman Series:
For the first line of the Lyman series:
\(
\begin{aligned}
& n_f=1 \\
& n_i=2
\end{aligned}
\)
Thus,
\(
\frac{1}{\lambda_1}=R \cdot 1^2\left(\frac{1}{1^2}-\frac{1}{2^2}\right)=R\left(1-\frac{1}{4}\right)=R \cdot \frac{3}{4}
\)
Therefore,\(\lambda_1=\frac{4}{3 R}\)
Calculate Wavelength for Balmer Series:
For the second line of the Balmer series:
\(
\begin{aligned}
n_f & =2 \\
n_i & =4
\end{aligned}
\)
Thus,
\(
\begin{gathered}
\frac{1}{\lambda_2}=R Z^2\left(\frac{1}{2^2}-\frac{1}{4^2}\right)=R Z^2\left(\frac{1}{4}-\frac{1}{16}\right) \\
=R Z^2\left(\frac{4-1}{16}\right)=R Z^2 \cdot \frac{3}{16}
\end{gathered}
\)
Therefore, \(\lambda_2=\frac{16}{3 R Z^2}\)
\(
\begin{aligned}
&\text { Since } \lambda_1=\lambda_2\\
&\frac{4}{3 R}=\frac{16}{3 R Z^2}
\end{aligned}
\)
\(
Z^2=\frac{16}{4}=4
\)
\(
Z=2
\)

Hint

(c) Use the photoelectric effect equation: \(E=\phi+e V_0\)
Substitute the given values: \(E=2.75 \mathrm{eV}+e(10 \mathrm{~V})\)
Simplify: \(E=2.75 \mathrm{eV}+10 \mathrm{eV}\)
Calculate: \({E}=12.75 \mathrm{eV}\)
Use the energy transition formula for hydrogen: \(E=13.6 \mathrm{eV}\left(\frac{1}{n_f^2}-\frac{1}{n_i^2}\right)\)
The electron transitions to the ground state, so \(n_f=1[latex] and [latex]n_i=n\).
Substitute the values: \(12.75 \mathrm{eV}=13.6 \mathrm{eV}\left(\frac{1}{1^2}-\frac{1}{n^2}\right)\)
\(
n^2=16
\)
\(
n=4
\)

Hint

(c) Energy levels of hydrogen atom: \(E_n=-\frac{13.6}{n^2} \mathrm{eV}\), where \(n\) is the principal quantum number ( \(n=1,2,3, \ldots\) ).
Energy of emitted photon: \(\Delta E=E_f-E_i\), where \(E_i\) and \(E_f\) are the initial and final energy levels.
\(
\begin{aligned}
&\text { Calculate the energy levels for } n=1,2,3,4 \text {. }\\
&\begin{aligned}
& E_1=-\frac{13.6}{1^2}=-13.6 \mathrm{eV} \\
& E_2=-\frac{13.6}{2^2}=-3.4 \mathrm{eV} \\
& E_3=-\frac{13.6}{3^2}=-1.51 \mathrm{eV} \\
& E_4=-\frac{13.6}{4^2}=-0.85 \mathrm{eV}
\end{aligned}
\end{aligned}
\)
\(
\begin{aligned}
& \Delta E_{21}=E_2-E_1=-3.4-(-13.6)=10.2 \mathrm{eV} \\
& \Delta E_{32}=E_3-E_2=-1.51-(-3.4)=1.89 \mathrm{eV} \approx 1.9 \mathrm{eV} \\
& \Delta E_{43}=E_4-E_3=-0.85-(-1.51)=0.66 \mathrm{eV} \approx 0.65 \mathrm{eV} \\
& \Delta E_{31}=E_3-E_1=-1.51-(-13.6)=12.09 \mathrm{eV} \\
& \Delta E_{42}=E_4-E_2=-0.85-(-3.4)=2.55 \mathrm{eV} \\
& \Delta E_{41}=E_4-E_1=-0.85-(-13.6)=12.75 \mathrm{eV}
\end{aligned}
\)
0.65 eV is possible \(\left(\Delta E_{43}\right)\).
1.9 eV is possible ( \(\Delta E_{32}\) ).
11.1 eV is not possible.
13.6 eV is possible ( \(E_1\) to \(n=\infty\) ).

Hint

(a) The energy of a hydrogen like atom in n-th orbit is given by
\(
E_n=-Z^2 \times \frac{13.6}{n^2} \mathrm{eV}
\)
For first excited state of \({H e}^{+}, n=2, Z=2\)
\(
\therefore E_{H e^{+}}=-\frac{4}{2^2} \times 13.6=-13.6 \mathrm{eV}
\)

Hint

(c) Kinetic energy of alpha particle: \(K E=\frac{1}{2} m v^2\)
Charge of the target nucleus: \(Z e\)
Charge of alpha particle: \(q_\alpha=2 e\)
At the distance of closest approach, the kinetic energy of the alpha particle is converted into potential energy.
Potential energy between two charges \(q_1\) and \(q_2\) separated by a distance \(r\) is given by \(P E=\frac{1}{4 \pi \epsilon_0} \frac{q_1 q_2}{r}\).
At the closest approach, the kinetic energy of the alpha particle is equal to the electrostatic potential energy:
\(K E=P E\)
\(\frac{1}{2} m v^2=\frac{1}{4 \pi \epsilon_0} \frac{(q_\alpha)(Z e)}{r}\)
\(
\begin{aligned}
& r=\frac{2}{4 \pi \varepsilon_0} \frac{q_\alpha Z e}{m v^2} \\
& \Rightarrow r \propto Z e \propto q_\alpha \propto \frac{1}{m} \propto \frac{1}{v^2}
\end{aligned}
\)

Hint

(c) Energy released when electron in the atom jumps from excited state \((n=3)\) to ground state \((n=1)\) is
\(
\begin{aligned}
& E=h v=E_3-E_1=\frac{-13.6}{3^2}-\left(\frac{-13.6}{1^2}\right) \\
& =\frac{-13.6}{9}+13.6=12.1 \mathrm{eV}
\end{aligned}
\)
Therefore, stopping potential
\(
\begin{aligned}
& e V_0=h v-\phi_0 \\
& =12.1-5.1\left[\because \text { work function } \phi_0=5.1\right] \\
& V_0=7 \mathrm{~V}
\end{aligned}
\)

Hint

(a) Given, Mass of Projectile \(=M_1\)
Mass of target nucleus \(=\mathrm{M}_2\)
Charge on Projectile particle, \(q_1=z_1\)
Charge on target nucleus, \(q_2=z_2\)
Then, the electrostatic potential energy between the two charges is given by
\(
U=\frac{1}{4 \pi \epsilon_0} \frac{q_1 q_2}{r}
\)
Distance of closest approach \(=r_0\)
Then, according to the definition of the distance of closest approach, at \(\mathrm{r}=\) \(r_0\), the kinetic energy becomes equal to coulombic potential i.e.
(K.E of the projectile particle) \()_{r=r_0}=(\) Potential energy between projectile and target nucleus) \()_{r=r_0}\)
or, \(\frac{1}{2} M_1 v^2=\frac{1}{4 \pi \epsilon_0} \frac{q_1 q_2}{r_0}\)
\(
\frac{1}{2} M_1 v^2=\frac{1}{4 \pi \epsilon_0} \frac{z_1 z_2}{r_0}
\)
i.e.Kinetic energy of the projectile \(\propto z_1 z_2\)

Hint

(c) Number of spectral lines emitted when an electron transitions from energy level \(n\) to lower levels is given by \(\frac{n(n-1)}{2}\).
Maximum wavelength corresponds to minimum energy difference between energy levels.

How to solve?
Determine the initial energy level \(n\) using the number of emitted wavelengths.
Identify the transition with the smallest energy difference.
Step 1: Determine the initial energy level \(n\)
The number of spectral lines is given by \(\frac{n(n-1)}{2}=6\).
Solving for \(n\) :
\(n(n-1)=12\)
\(n^2-n-12=0\)
\((n-4)(n+3)=0\)
\(n=4\) (since \(n\) must be positive)
Thus, the electron is initially at energy level \(n=4\).
Step 2: Identify the transition with the smallest energy difference
The smallest energy difference corresponds to the transition between adjacent energy levels.
In this case, it’s the transition from \(n=4\) to \(n=3\).

Solution: The maximum wavelength of emitted radiation corresponds to the transition from \(n=4\) to \(n=3\).

Hint

(c) When the electron is in first excited state ( \(\{n}=2\) ), the excitation energy is given by

\(
\begin{aligned}
&\Delta \mathrm{E}=\mathrm{E}_2-\mathrm{E}_1\\
&\text { We have, } E_n=-\frac{13.6}{n^2} e V
\end{aligned}
\)
\(
\begin{aligned}
& \therefore E_2=-\frac{13.6}{2^2} \mathrm{eV}=-3.4 \mathrm{eV} \\
& \text { Given } E_1=-13.6 \mathrm{eV} \\
& \therefore \Delta \mathrm{E}=(-3.4)-(-13.6)=10.2 \mathrm{eV}
\end{aligned}
\)

Hint

(a)

\(
\begin{aligned}
&\text { (d) Energy in the first excited state }\\
&\begin{aligned}
& =\frac{-13.6}{n^2}=\frac{-13.6}{2^2}=-3.4 \mathrm{eV} \\
& \text { But K.E. }=-(\text { Total energy })=+3.4 \mathrm{eV} .
\end{aligned}
\end{aligned}
\)

Hint

(a) Radius of a nucleus is given by:
\(
r \propto A^{1 / 3}
\)
where \(r\) is the radius and \(A\) is the mass number of the nucleus.
Using the proportionality, we can express the radii of Aluminum and Tellurium as:
\(
\frac{r_{\mathrm{Al}}}{r_{\mathrm{Te}}}=\frac{A_{\mathrm{Al}}^{1 / 3}}{A_{\mathrm{Te}}^{1 / 3}}
\)
The mass number of Aluminum \(A_{\mathrm{Al}}=27\) and the mass number of Tellurium \(A_{\mathrm{Te}}=125\). Therefore, we can write:
\(
\frac{r_{\mathrm{Al}}}{r_{\mathrm{Te}}}=\frac{27^{1 / 3}}{125^{1 / 3}}
\)
\(
r_{\mathrm{Te}}=r_{\mathrm{Al}} \cdot \frac{5}{3}
\)
\(
\begin{aligned}
&\text { Given that } r_{\mathrm{Al}}=3.6 \mathrm{fm} \text { : }\\
&r_{\mathrm{Te}}=3.6 \mathrm{fm} \cdot \frac{5}{3}=6 \mathrm{fm}
\end{aligned}
\)

Hint

(b) Energy levels of hydrogen atom: \(E_n=-\frac{13.6}{n^2} \mathrm{eV}\), where \(n\) is the principal quantum number.
Number of spectral lines emitted when an electron transitions from level \(n\) to lower levels: \(\frac{n(n-1)}{2}\).
Step 1: Calculate the excited energy level
The energy of the electron in the ground state \((n=1)\) is \(E_1=-13.6 \mathrm{eV}\).
After absorbing a photon of 12.1 eV, the electron’s energy becomes
\(
E=-13.6 \mathrm{eV}+12.1 \mathrm{eV}=-1.5 \mathrm{eV}
\)
We have \(E_n=\frac{-13.6}{n^2} \mathrm{eV}=-1.5 \mathrm{eV}\).
Solving for \(n\) :
\(n^2=\frac{-13.6}{-1.5} \approx 9\)
\(n=3\)
The electron is excited to the \(n=3\) level.
Step 2: Calculate the number of spectral lines
The number of spectral lines emitted when an electron transitions from level \(n[latex] to lower levels is given by [latex]\frac{n(n-1)}{2}\).
For \(n=3\), the number of spectral lines is \(\frac{3(3-1)}{2}=\frac{3 \times 2}{2}=3\).

Hint

(d) Let the radius of \({ }_4^9 \mathrm{Be}\) nucleus be \(r\). Then, radius of germanium ( \(G e\) ) nucleus will be \(2 r\). Radius of nucleus is given by
\(
\begin{aligned}
& R=R_0 A^{1 / 3} \\
& \therefore \frac{R_1}{R_2}=\left(\frac{A_1}{A_2}\right)^{1 / 3} \\
& \Rightarrow \frac{r}{2 r}=\left(\frac{9}{A_2}\right)^{1 / 3}\left(\because A_1=9\right) \\
& \Rightarrow\left(\frac{1}{2}\right)^3=\frac{9}{A_2}
\end{aligned}
\)
Hence, \(A_2=9 \times(2)^3=9 \times 8=72\)
Thus, in germanium (Ge) nucleus number of nucleons is 72.

Hint

(d) Let the energy in \(\mathrm{A}, \mathrm{B}\) and C state be \(E_A, E_B\) and \(E_C\), then from the figure

\(
\begin{aligned}
& \left(E_C-E_B\right)+\left(E_B-E_A\right)=\left(E_C-E_A\right) \text { or } \frac{h c}{\lambda_1}+\frac{h c}{\lambda_2}=\frac{h c}{\lambda_3} \\
& \Rightarrow \lambda_3=\frac{\lambda_1 \lambda_2}{\lambda_1+\lambda_2}
\end{aligned}
\)

Hint

(b) Total energy of the electron: \(E=-3.4 \mathrm{eV}\)
The relationship between total energy ( \({E}\) ) and kinetic energy ( \({K}\) ) for an electron in an atom is: \(K=-E\).

How to solve?
Use the relationship between total energy and kinetic energy to find the kinetic energy.
Step 1: Calculate the kinetic energy
Use the formula \(K=-E\).
Substitute the given value of \(E=-3.4 \mathrm{eV}\).
\(K=-(-3.4 \mathrm{eV})\)
\(K=3.4 \mathrm{eV}\)

Solution: The kinetic energy of the electron is 3.4 eV.

Hint

(b) In Bohr’s model, angular momentum is quantised i.e \(L=n\left(\frac{h}{2 \pi}\right)\)

Explanation: 

Bohr’s model: introduced the idea that electrons orbit the nucleus in specific, quantized energy levels. It postulates that electrons can only have certain allowed angular momenta. This quantization of angular momentum is a key concept in Bohr’s model, explaining the stability of the atom and the observed line spectra of hydrogen.

Hint

(c) Energy of hydrogen atom: \(E=-\frac{13.6}{n^2} \mathrm{eV}\)
Initial state: \({n}_{{i}}=3\)
Final state: \(n_f=2\)
\(
\begin{aligned}
& \Delta E=E_f-E_i \\
& \Delta E=-\frac{13.6}{4} \mathrm{eV}-\left(-\frac{13.6}{9} \mathrm{eV}\right)
\end{aligned}
\)
\(
\Delta E \approx 1.89 \mathrm{eV}
\)
The energy of the emitted photon is approximately 1.9 eV.

Hint

(a) The principal quantum number \(n=1\).
Four systems: doubly ionized lithium \(\left({L i^{2+}}^{2+}\right)\), singly ionized helium \(\left(H e^{+}\right)\), deuterium atom ( \(D\) ), and hydrogen atom ( \(\boldsymbol{H}\) ).

The radius of the \(n\)-th orbit in a hydrogen-like atom is given by \(r_n=\frac{a_0 n^2}{Z}\), where \(a_0\) is the Bohr radius and \(Z\) is the atomic number.
For the first orbit \((n=1)\), the radius is \(r_1=\frac{a_0}{Z}\).

How to solve
Compare the atomic numbers of the given systems and find the one with the highest atomic number, which will have the smallest radius.
Step 1:
Determine the atomic number \((Z)\) for each system.
For doubly ionized lithium \(\left(L i^{2+}\right), Z=3\).
For singly ionized helium ( \(\mathrm{He}^{+}\)), \({Z}=2\).
For deuterium atom ( \(D[latex] ), [latex]Z=1\).
For hydrogen atom \(({H}), Z=1\).
Step 2:
Calculate the radius of the first orbit ( \(n=1\) ) for each system.
For \(L i^{2+}, r_1=\frac{a_0}{3}\).
For \(H e^{+}, r_1=\frac{a_0}{2}\).
For \(D, r_1=\frac{a_0}{1}=a_0\).
For \(H_{,} r_1=\frac{a_0}{1}=a_0\).
Step 3:
Compare the radii.
The smallest radius corresponds to the largest \(Z\).
Since \(3>2>1\), the smallest radius is for \(\mathrm{Li}^{2+}\).

Solution: The system with the minimum radius of the first orbit is doubly ionized lithium.

Hint

(d) In the question, an electron is revolving around the nucleus, this means that one charge is an electron, and the other one is a proton. We know that the charge on the electron is one unit negative so it can be written as \(-e\) and the charge on the proton is one unit positive so it can be written as \(+e\), and the radius can be denoted as \(r\). So, we can write the formula as:
\(
F \propto \frac{(-e)(+e)}{r^2}
\)
Now, removing the proportionality constant we can introduce a constant \(k\). This will be:
\(
F=-k \frac{e^2}{r^2}
\)
Now, we have define the direction of the forces and the direction will be along the radius that is written as \(\hat{r}\)
This \(\hat{r}\) is equal to:
\(
\hat{r}=\frac{\vec{r}}{r}
\)
Now, multiplying this direction of force into the value of force, we get:
\(
\begin{aligned}
F & =-k \frac{e^2}{r^2} \times \frac{\vec{r}}{r} \\
F & =-k \frac{e^2}{r^3} \vec{r}
\end{aligned}
\)

Hint

(d) \(J J\) Thomson’s cathode-ray tube experiment demonstrated the relation for \(e / m\) of charged particles. The relation is
\(
\frac{e}{m}=\frac{E^2}{2 B^2 V}
\)
Thus, knowing \(E, B\) and \(V\), the value of \(e / m\) for electrons and for protons can be calculated. It is found the \(e / m\) of electrons is much greater than the \(e / m\) of protons.

Hint

(a) We have \(E_n=\frac{-2 \pi^2 m K^2 Z^2 e^4}{n^2 h^2}\).
\(
E_n \propto Z^2
\)
Given, \(Z_H=1, Z_{H e}=2\)
\(
\begin{aligned}
& \therefore \frac{E_H}{E_{H e}}=\frac{Z_H^2}{Z_{H e}^2}=\frac{1}{4} \\
& \Rightarrow E_{H e}=4 E_H
\end{aligned}
\)
Given, \(E_H=E_n\)
\(
\therefore E_{H e}=4 E_n
\)

Hint

(b) Wavelength of lines emitted is \(\frac{1}{\lambda}=R\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)\) where \(R\) Rydbergs constant.
Given, \(n_1=2, n_2=4 \therefore \frac{1}{\lambda}=R\left(\frac{1}{2^2}-\frac{1}{4^2}\right)=\frac{3 R}{16} \Rightarrow \lambda=\frac{16}{3 R}\)

Hint

(c)  When an electron jumps from the fourth orbit to the second orbit, the emitted radiation corresponds to the second line of the Balmer series.

Explanation:
Balmer series:
This series refers to spectral lines in hydrogen where the electron transitions from a higher energy level ( \(n>2\) ) to the second energy level \((n=2)\).

Specific Transition:
An electron jumping from the fourth orbit ( \(n=4\) ) to the second orbit ( \(n=2\) ) produces a spectral line that is part of the Balmer series. This specific transition is known as the second line of the Balmer series.

Lyman series:
This series arises from transitions where the electron falls to the first energy level \(({n}=1)\).

Paschen series:
This series arises from transitions where the electron falls to the third energy level ( \({n}=3\) ).

Pfund series:
This series arises from transitions where the electron falls to the fourth energy level \((n=4)\).

Hint

(a) Atomic hydrogen is unstable and it has life period of a fraction of a second.

Hint

(a) Millikan determined the mass of the electron by measuring the charge of the electron.

Hint

(a) Whenever an atom gets de-excited from higher to the lower orbit, it radiation of a given frequency
\(
\Delta E=h f \Rightarrow f \propto \Delta E
\)
The photons of the highest frequency will be emitted, for the transition in which \(\Delta E\) is maximum, i.e., from 2 to 1.

Note: \(
\text { Frequency, } \nu=R c\left[\frac{1}{n_1^2}-\frac{1}{n_2^2}\right]
\)

Hint

(a) Coulomb’s force formula: \(F=\frac{1}{4 \pi \varepsilon_0} \frac{q_1 q_2}{r^2}\)
Centripetal force formula: \(F=\frac{m v^2}{r}\)

How to solve
Equate the Coulomb force to the centripetal force and solve for \(v\).
Step 1:
Equate Coulomb force and centripetal force
The Coulomb force between the electron and proton is:
\(F_c=\frac{1}{4 \pi \varepsilon_0} \frac{e^2}{r^2}\)
The centripetal force acting on the electron is:
\(F_{c p}=\frac{m v^2}{r}\)
Equating the two forces:
\(\frac{1}{4 \pi \varepsilon_0} \frac{e^2}{r^2}=\frac{m v^2}{r}\)
Step 2:
Solve for \(v\)
Multiply both sides by \(r\) :
\(\frac{1}{4 \pi \varepsilon_0} \frac{e^2}{r}=m v^2\)
Divide both sides by \(m\) :
\(v^2=\frac{1}{4 \pi \varepsilon_0} \frac{e^2}{m r}\)
Take the square root of both sides:
\(v=\sqrt{\frac{1}{4 \pi \varepsilon_0} \frac{e^2}{m r}}\)
Simplify:
\(v=\frac{e}{\sqrt{4 \pi \varepsilon_0 m r}}\)

Solution: The velocity of the electron is \(\frac{e}{\sqrt{4 \pi m \varepsilon_0 r}}\).

Hint

(a) The radius of the \(n\)-th Bohr orbit is given by:
\(r_n=r_0 n^2\)

For the first excited state, \(n=2\).
Substituting \(n=2\) into the formula:
\(r_2=r_0(2)^2\)
\(r_2=4 r_0\)
Therefore, the radius of the hydrogen atom in the first excited state is four times the Bohr radius.

Hint

(c)

Explanation:
Sodium vapor lamps emit light of specific wavelengths, resulting in a characteristic emission spectrum. This is because excited sodium atoms release energy in the form of photons of those wavelengths, creating a pattern of bright lines in the spectrum.

Why other options are incorrect:
(a) band spectrum:

Band spectra are produced by molecules, not individual atoms like in a sodium vapor lamp.
(b) continuous spectrum:

A continuous spectrum contains all wavelengths of light, which is not the case for sodium vapor lamps.
(d) absorption spectrum:

Absorption spectra show dark lines where specific wavelengths have been absorbed, which is the opposite of what a sodium vapor lamp emits.

Hint

(b) P.E. \(=-\frac{K Z e^2}{r}\) and \(K \cdot E \cdot=\frac{K Z e^2}{2 r}\),
where, \(r\) is the radius of orbit which increases as we move from ground to an excited state.
Therefore, when a hydrogen atom is raised from the ground state, it increases the value of \(r\).
As a result of this, P.E. increases (decreases in negative) and K.E. decreases.

Hint

\(
\begin{aligned}
& \text { (b) } r \propto n^2 \\
& \therefore \frac{\text { radius of final state }}{\text { radius of initial state }}=n^2 \\
& \frac{21.2 \times 10^{-11}}{5.3 \times 10^{-11}}=n^2 \\
& \therefore n^2=4 \text { or } n=2
\end{aligned}
\)

Hint

(d) The relationship between the impact parameter \(b\) and the scattering angle \(\theta\) is given by:
\(b=\frac{k}{K E} \cot \frac{\theta}{2}\)
Where \(k\) is a constant, and \(K E\) is the kinetic energy of the alpha particle.
Substitute \(b=0\) into the formula:
\(
0=\frac{k}{K E} \cot \frac{\theta}{2}
\)
For the equation to hold true, \(\cot \frac{\theta}{2}\) must be equal to 0 :
\(\cot \frac{\theta}{2}=0\)
The cotangent function is zero at \(\frac{\pi}{2}\) or \(90^{\circ}\) :
\(\frac{\theta}{2}=\frac{\pi}{2}\)
Solve for \(\theta\) :
\(\theta=\pi\)
\(\theta=180^{\circ}\)
The scattering angle for an impact parameter of zero is \(180^{\circ}\).

Hint

(b) Neon street sign. Neon signs produce a line emission spectrum because they are a type of gas-discharge lamp that emits light when an electric current passes through a gas (like neon) at low pressure, causing the atoms to become excited and release specific wavelengths of light.

Explanation:
Line emission spectrum:
A line spectrum is a spectrum of light that consists of bright lines at specific wavelengths, characteristic of the element producing the light.

Neon signs:
Neon signs are made of glass tubes filled with neon or other gases. When an electric current is passed through the gas, the atoms become excited and emit light of specific wavelengths, resulting in the characteristic colored glow.

Electric fire:
While electric fires produce light, they also produce a continuous spectrum of light, not just lines.

Red traffic light:
Red traffic lights are also not a source of a line emission spectrum. They use specific colored materials, but also are not gas-discharge lamps.

Sun:
The Sun emits a continuous spectrum of light because it is a hot, dense object that produces light across all wavelengths.

Hint

(a) As \(r \propto n^2\), therefore, radius of 2nd Bohr’s orbit \(=4 a_0\).

Hint

\(
\begin{aligned}
& \text { (d) } E=E_4-E_3 \\
& =-\frac{13.6}{4^2}-\left(-\frac{13.6}{3^2}\right)=-0.85+1.51 \\
& =0.66 \mathrm{eV}
\end{aligned}
\)

Hint

\(
\begin{aligned}
&\text { (a) Second excited state corresponds to } n=3\\
&\therefore \quad E=\frac{13.6}{3^2} \mathrm{eV}=1.51 \mathrm{eV}
\end{aligned}
\)

Hint

(b) To explain his theory, Bohr used the conservation of angular momentum. He proposed that the angular momentum of an electron in a given orbit is quantized and conserved. This concept helps to explain why electrons orbit the nucleus without spiraling into it.

Bohr used quantisation of angular momentum.
For stationary orbits, Angular momentum \(I \omega=\frac{n h}{2 \pi}\) where \(n=1,2,3, \ldots\) etc

Hint

(d) The ionization energy \((E)\) is proportional to the square of the atomic number \((Z)\) :
\(E \propto Z^2\)

Therefore, the ratio of the ionization energy of helium \(\left(E_{H e}\right)\) to that of hydrogen \(\left(E_H\right)\) is:
\(\frac{E_{H e}}{E_H}=\frac{Z_{H e}^2}{Z_U^2}\)
Substitute the atomic numbers of helium ( \(Z_{H e}=2\) ) and hydrogen \(\left(Z_H=1\right)\) into the equation:
\(
\frac{E_{H e}}{E_H}=\frac{2^2}{1^2}=\frac{4}{1}=4
\)
Multiply the ionization energy of hydrogen by the ratio to find the ionization energy of helium:
\(E_{H e}=4 \times E_H\)
\(E_{H e}=4 \times 13.6 \mathrm{eV}\)
\(E_{H e}=54.4 \mathrm{eV}\)
The ionization energy of a helium atom is 54.4 eV.

Hint

(a) 53 electrons in iodine atom are distributed as \(2,8,18,18,7\)
\(
\begin{aligned}
\therefore n & =5 \\
r_n & =\left(0.53 \times 10^{-10}\right) \frac{n^2}{Z} \\
& =\frac{0.53 \times 10^{-10} \times 5^2}{53}=2.5 \times 10^{-11} \mathrm{~m}
\end{aligned}
\)