Atomic & Molecular Physics MCQ Quiz in বাংলা - Objective Question with Answer for Atomic & Molecular Physics - বিনামূল্যে ডাউনলোড করুন [PDF]

Concept:

An electric dipole transition is the dominant effect of an interaction of an electron in an atom with the electromagnetic field.

Calculation:

For, dipole transition,

Δl = ± 1 and Δ m = 0, ± 1

For all options, n = 2, so l = 0,1

For l = 0, m = 0 and for l, m = -1, 0, 1

The transitions |3, 0, 0〉 → |2, 1, 0〉 → |1, 0, 0〉 for m = -1, 0, 1 are all valid according to the dipole transition rule.

Thus, there are three different states of equal probability.

Hence each transition has a probability of 1/3.

The correct answer is option (3).

Explanation:

The splitting of spectral lines in the normal Zeeman effect is given by the formula:

\(\Delta=\mu_BB\Delta m\)

which translates into wavelength splitting as:

\(\Delta \lambda=\frac{eB\lambda^2}{4\pi m_e C}\)

From the equation, we see that splitting is directly proportional to the charge-to-mass ratio e/me​.

• If $e/m_e$  increases, then splitting increases.
• If $e/m_e$​ decreases, then splitting decreases.
• If $e/m_e$ remains the same, splitting remains the same.
• The material of the light source does not affect the normal Zeeman effect since it depends only on fundamental constants.

Thus, option '1' is correct: Increase proportionally (if $e/m_e$ increases).

Concept:

Energy Difference in Rotational States:

• In a diatomic molecule like HD, the energy difference \( \Delta E \) between the two lowest rotational states is determined by the formula:
• \(\Delta E = \frac{h^2}{8 \pi^2 I} = \frac{h^2}{8 \pi^2 \mu r^2}\)
• Where \(h\) is Planck's constant, \( \mu \) is the reduced mass, and \( r \) is the mean distance between the atoms.
• The reduced mass \( \mu \) for the HD molecule is given by:
• \(\mu = \frac{m_H m_D}{m_H + m_D}\) , where m_H is the mass of hydrogen and m_D is the mass of deuterium.
• Substitute the given values into the formula to calculate the energy difference.

Given data:

• Distance between atoms, \(r = 0.08 \, \text{nm} = 8 \times 10^{-11} \, \text{m}\)
• Mass of hydrogen atom, \(m_H = 938 \, \text{MeV}/c^2\)
• Mass of deuterium atom, \(m_D = 1876 \, \text{MeV}/c^2\)

Now, the energy difference between the two lowest rotational states is approximately:

\( \Delta E \approx 10^{-2} \, \text{eV} \)

Solution:

Hence, the correct answer is:

• Option 2: \(10^{-2} \, \text{eV}\)

Concept:

Spin-Orbit Interaction and Energy Splitting:

• The spin-orbit interaction is the interaction between the electron's spin angular momentum and its orbital angular momentum. This interaction leads to the splitting of energy levels in atoms.
• The Hamiltonian for spin-orbit interaction is given by: \( H = a \, \vec{L} \cdot \vec{S} \)
• The energy splitting between levels is determined by the quantum numbers associated with orbital and spin angular momenta: \( L \) and \( S \).

For the levels \( 2P_{3/2} \) and \( 2P_{1/2} \), the energy splitting is given by:

Energy splitting = \( \Delta E = a \cdot (L + S) \)

The correct expression for the splitting between the \( 2P_{3/2} \) and \( 2P_{1/2} \) levels is \( \frac{3}{2} a h^2 \).

Solution:

Hence, the correct answer is:

• Option 1: \( \frac{3}{2} a h^2 \)

Concept:

Spectral Terms and Electronic Configuration:

• The spectral terms arise from the combination of the quantum numbers associated with the total orbital angular momentum (L) and total spin angular momentum (S) of electrons.
• The term with the lowest energy corresponds to the highest spin multiplicity and highest L value (for terms with the same L value, the term with the highest spin multiplicity is the lowest in energy).

Term Representation: The spectral terms are represented as ^2S+1L, where S is the total spin quantum number, L is the total orbital angular momentum quantum number, and 2S+1 gives the spin multiplicity.

Solution:

The given spectral terms are: ³D, ¹D, ³P, ⁵S, and ³S.

Now, considering the spin multiplicity and total orbital angular momentum quantum numbers:

• For the term ³D: S = 1, L = 2, and the spin multiplicity = 2S + 1 = 3
• For the term ¹D: S = 0, L = 2, and the spin multiplicity = 2S + 1 = 1
• For the term ³P: S = 1, L = 1, and the spin multiplicity = 2S + 1 = 3
• For the term ⁵S: S = 2, L = 0, and the spin multiplicity = 2S + 1 = 5
• For the term ³S: S = 1, L = 0, and the spin multiplicity = 2S + 1 = 3

The term with the highest spin multiplicity and the lowest total orbital angular momentum quantum number corresponds to the lowest energy.

The correct answer is the term with the highest spin multiplicity: ⁵S.

The correct option is: Option 1 (⁵S).

Concept:

Hydrogen Spectrum:

• The wavelength of spectral lines in the hydrogen spectrum can be calculated using the Rydberg formula:

\( \frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \)

• R : Rydberg constant.
• n₁ and n₂ : Lower and higher energy levels, respectively, where \(n_2 > n_1\) .

Balmer Series: Occurs when\( n_1 = 2\) and \( n_2 = 3, 4, 5, \dots .\)

Lyman Series: Occurs when n₁ = 1 and \(n_2 = 2, 3, 4, \dots \) .

Solution:

If the wavelength of the first line of the Balmer series is \lambda , we use the Rydberg formula for Balmer's first line ( \(n_1 = 2, n_2 = 3\) ):

• \( \frac{1}{\lambda} = R \left( \frac{1}{2^2} - \frac{1}{3^2} \right) = R \left( \frac{1}{4} - \frac{1}{9} \right) \)
• \( \frac{1}{\lambda} = R \left( \frac{5}{36} \right) \).

For the first line of the Lyman series ( \(n_1 = 1, n_2 = 2\) ):

• \( \frac{1}{\lambda_{\text{Lyman}}} = R \left( \frac{1}{1^2} - \frac{1}{2^2} \right) = R \left( 1 - \frac{1}{4} \right) \)
• \( \frac{1}{\lambda_{\text{Lyman}}} = R \left( \frac{3}{4} \right) \).

The ratio of the wavelengths:

• \( \frac{\lambda_{\text{Lyman}}}{\lambda} = \frac{\frac{36}{5R}}{\frac{4}{3R}} = \frac{36}{5} \times \frac{3}{4} = \frac{27}{5} \).
• \( \lambda_{\text{Lyman}} = \frac{5}{27} \lambda \).

The correct option is Option 2.

Concept:

Hyperfine Splitting:

• Hyperfine splitting arises from the interaction between the electronic angular momentum (\( J \)) and the nuclear spin (\( I \)).
• The total angular momentum quantum number \( F \) is defined as:
  • \( F = |J - I|, |J - I| + 1, \dots, J + I \)
• The number of hyperfine levels is determined by the possible values of \( F \) for each state.

Solution:

For the given transition \( ^2P_{1/2} \to ^2S_{1/2} \):

• The electronic angular momentum quantum number for both levels is \( J = \frac{1}{2} \).
• The nuclear spin quantum number is given as \( I = \frac{1}{2} \).

The number of hyperfine components for each state is determined by the possible values of \( F \):

• For both \( J = \frac{1}{2} \) and \( I = \frac{1}{2} \), the possible values of \( F \) are:
  • \( F = 0, 1 \) (two levels for each state).

The total number of hyperfine components observed in the transition is:

• \( 2 \times 2 = 4 \)

The correct option is Option 2.

Concept:

Total Angular Momentum in L-S Coupling:

• In L-S coupling, the total angular momentum quantum number (\( J \)) is determined by combining the individual orbital angular momentum quantum numbers \( l_1 \) and \( l_2 \).
• The possible values of \( J \) are given by the rule:
  • \( J = |l_1 - l_2|, |l_1 - l_2| + 1, \dots, l_1 + l_2 \)

Solution:

The orbital angular momentum quantum numbers are:

• \( l_1 = 1 \)
• \( l_2 = 2 \)

Using the rule \( J = |l_1 - l_2|, |l_1 - l_2| + 1, \dots, l_1 + l_2 \), the possible values of \( J \) are:

• \( J = |1 - 2| = 1, 2, 3 \)

The correct option is Option 2.

Concept:

Selection Rules for Electric Dipole Transitions:

• The selection rules describe the allowed changes in quantum numbers during electronic transitions.
• For electric dipole transitions:
  • Change in orbital angular momentum quantum number (\(\Delta L\)): \(\Delta L = 0, \pm 1\) (but \(\Delta L = 0\) is forbidden for transitions between the same orbital).
  • Change in spin angular momentum (\(\Delta S\)): \(\Delta S = 0\) (spin must remain constant).
  • Change in total angular momentum quantum number (\(\Delta J\)): \(\Delta J = 0, \pm 1\) (except \(\Delta J = 0\) for transitions between states with \(J = 0\)).

Solution:

The selection rules for electric dipole transitions are:

• \(\Delta L = 0, \pm 1\): The orbital angular momentum can change by 0 or ±1.
• \(\Delta S = 0\): Spin angular momentum must remain constant.
• \(\Delta J = 0, \pm 1\): The total angular momentum quantum number can change by 0 or ±1, but \(\Delta J = 0\) is forbidden if \(J = 0\).

The given rules match Option 3, where:

• \(\Delta L = 0, \pm 1\)
• \(\Delta S = 0 \)
• \(\Delta J = \pm 1\)

Correct Option: Option 3

Concept:

The positronium atom is a bound system consisting of an electron and a positron. Since the masses of the electron ( m_e ) and positron ( m_p ) are identical, the reduced mass ( \(μ\) ) of the positronium system is given as:

\(μ = \frac{m_e \cdot m_p}{m_e + m_p} = \frac{m_e}{2}\)

For a hydrogen atom, the reduced mass is approximately equal to the mass of the electron ( m_e ) because the proton's mass is significantly larger than the electron's.

The energy ( E ) and radius ( r ) of the nth orbit in a Bohr atom are inversely and directly proportional to the reduced mass ( μ)  respectively:

• Energy:\( E \propto -\frac{μ}{n^2}\)
• Radius:\( r \propto \frac{1}{μ}\)

Explanation:

• For the positronium atom, since the reduced mass ( \(μ\) ) is halved compared to the hydrogen atom, the radius ( \(r_p \) ) doubles and the energy ( E_p ) is halved.
• Thus, the radius of positronium ( r_p ) is 2r_H , and the energy ( E_p ) is E_H/2 .

Correct Answer: Option 1