Atomic & Molecular Physics MCQ Quiz - Objective Question with Answer for Atomic & Molecular Physics - Download Free PDF

Last updated on Jun 25, 2025

Latest Atomic & Molecular Physics MCQ Objective Questions

Atomic & Molecular Physics Question 1:

Consider the Bromine ion Br+in its ground state. The atomic number of Br is 35. The fine structure term symbol (2S+1𝐿j under the 𝐿𝑆 coupling scheme for the lowest energy state of this ion would be 

Answer (Detailed Solution Below)

Option 1 :

Atomic & Molecular Physics Question 1 Detailed Solution

Calculation:

Br+ has electronic configuration: [Ar] 4s2 3d10 4p4. Optical active electrons are equivalent electrons. The p4 configuration has exactly the same spectral representation as p2, but p4 has inverted order of energy levels.

This means: highest S corresponds to lowest energy, highest L corresponds to lowest energy, and highest J corresponds to lowest energy.

So, 3P2 has the lowest energy (here S and D have singlet state).

Atomic & Molecular Physics Question 2:

In a laboratory experiment on the normal Zeeman effect, a magnetic field of 0.5 T is applied to a light source emitting at 600 nm. How would the spectral line splitting change if the electron charge-to-mass ratio were slightly different from its accepted value?

  1.  Increase proportionally
  2. Decrease proportionally
  3.  Remain the same
  4. Depend on the light source material

Answer (Detailed Solution Below)

Option 1 :  Increase proportionally

Atomic & Molecular Physics Question 2 Detailed Solution

Explanation:

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

which translates into wavelength splitting as:

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

  • If e/mee/m_e  increases, then splitting increases.
  • If e/mee/m_e decreases, then splitting decreases.
  • If e/mee/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/mee/m_e increases).

Atomic & Molecular Physics Question 3:

 Consider the hydrogen-deuterium molecule HD. If the mean distance between the two atoms is 0.08 nm and the mass of the hydrogen atom is 938 MeV/c2, then the energy difference ΔE between the two lowest rotational states is approximately  

  1. 10-1eV
  2. 10-2​eV
  3. 2× 10-4​eV
  4. 10-3​eV

Answer (Detailed Solution Below)

Option 2 : 10-2​eV

Atomic & Molecular Physics Question 3 Detailed Solution

Concept:

Energy Difference in Rotational States:

  • In a diatomic molecule like HD, the energy difference between the two lowest rotational states is determined by the formula:
  • Where is Planck's constant, is the reduced mass, and is the mean distance between the atoms.
  • The reduced mass for the HD molecule is given by:
  • , where mH is the mass of hydrogen and mD is the mass of deuterium.
  • Substitute the given values into the formula to calculate the energy difference.

Given data:

  • Distance between atoms,
  • Mass of hydrogen atom,
  • Mass of deuterium atom,

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

Solution:

Hence, the correct answer is:

  • Option 2:

Atomic & Molecular Physics Question 4:

The spin-orbit interaction in an atom is given by H = a L.S, where L and S denote the orbital and spin angular momenta, respectively, of the electron. The splitting between the levels 2P3/2 and 2P1/2 is 

  1. (3/2)aℏ2 
  2. (1/2)aℏ2 
  3. (3)aℏ2 
  4. (5/2)aℏ2 

Answer (Detailed Solution Below)

Option 1 : (3/2)aℏ2 

Atomic & Molecular Physics Question 4 Detailed Solution

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:
  • The energy splitting between levels is determined by the quantum numbers associated with orbital and spin angular momenta: .

For the levels and , the energy splitting is given by:

Energy splitting =

The correct expression for the splitting between the and levels is .

Solution:

Hence, the correct answer is:

  • Option 1:

Atomic & Molecular Physics Question 5:

The first absorption spectrum of 12C16O is at 3.842 cm-1 while that of 13C16O is at 3.673 cm-1 . The ratio of their moments of inertia is  

  1. 1.851
  2. 1.286
  3. 1.046
  4. 1.038

Answer (Detailed Solution Below)

Option 3 : 1.046

Atomic & Molecular Physics Question 5 Detailed Solution

Concept:

Absorption Spectrum and Moment of Inertia:

  • The frequency of absorption ( ν  ) for a rotating molecule is related to its moment of inertia ( I ) by the formula: .
  • Here, h is Planck's constant, c is the speed of light, and I is the moment of inertia.
  • The absorption frequency ( ν ) is inversely proportional to the moment of inertia ( I ), so we can calculate the ratio of moments of inertia by taking the ratio of the frequencies.

Given:

  • ν' for

The ratio of moments of inertia can be calculated as:

Thus, the ratio of moments of inertia is approximately 1.046 when considering the reverse ratio.

Solution:

Hence, the correct answer is:

  • Option 3: 1.046

Top Atomic & Molecular Physics MCQ Objective Questions

The energy (in keV) and spin-parity values E (JP) of the low lying excited states of a nucleus of mass number A = 152, are 122(2+), 366(4+), 707 (6+), and 1125(8+). It may be inferred that these energy levels correspond to a

  1. rotational spectrum of a deformed nucleus
  2. rotational spectrum of a spherically symmetric nucleus
  3. vibrational spectrum of a deformed nucleus
  4. vibrational spectrum of a spherically symmetric nucleus

Answer (Detailed Solution Below)

Option 1 : rotational spectrum of a deformed nucleus

Atomic & Molecular Physics Question 6 Detailed Solution

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Explanation:

  • The given energy levels and their respective spin-parity values indicate a 'rotational spectrum' of a 'deformed nucleus'.
  • The fact that the energy levels are increasing but not uniformly, and considering the spin and parity, strongly suggests a pattern consistent with a nuclear deformation leading to a rotational energy level scheme.
  • Furthermore, the parity remains positive across transitions, following the (2n)+ rule typical for even-even deformed nuclei. So together, these characteristics point to an even-even deformed nucleus exhibiting a rotational spectrum.

The mean kinetic energy per atom in a sodium vapour lamp is 0.33 eV. Given that the mass of sodium atom is approximately 22.5 × 109eV, the ratio of the Doppler width of an optical line to its central frequency is

  1. 7 × 10-7
  2. 6 × 10-6
  3. 5 × 10-5
  4. 4 × 10-4

Answer (Detailed Solution Below)

Option 2 : 6 × 10-6

Atomic & Molecular Physics Question 7 Detailed Solution

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CONCEPT:

Doppler broadening is the broadening of spectral lines due to the Doppler effect caused by a distribution of velocities of atoms or molecules. Different velocities of the emitting (or absorbing) particles result in different Doppler shifts, the cumulative effect of which is the emission (absorption) line broadening.

Mathematically, the Doppler width of an optical line Δ ν0  = 

Where, ν0 = central frequency, kB = Boltzmann's constant, T = temperature

m = mass and c = velocity of light in vacuum

EXPLANATION:

Doppler shift is: 

⇒ Δ ν0  = 

Hence the correct answer is option 2.

The red line of wavelength 644 nm in the emission spectrum of Cd corresponds to a transition from the 1D2 level to the 1P1 level. In the presence of a weak magnetic field, this spectral line will split into (ignore hyperfine structure)

  1. 9 lines
  2. 6 lines
  3. 3 lines
  4. 2 lines

Answer (Detailed Solution Below)

Option 3 : 3 lines

Atomic & Molecular Physics Question 8 Detailed Solution

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Explanation:

If only the normal or "weak field" Zeeman effect is being considered, where splitting is independent of the number of available orbitals and only dependent on the change in the magnetic quantum number (m = -1, 0, +1), then the transition would indeed split into three lines.

This occurs because the orbital angular momentum can adjust to the magnetic field, and the result is three transitions with Δm = -1, 0, +1 respectively that correspond to the three spectral lines

The outermost shell of an atom of an element is 3d3. The spectral symbol for the ground state is

  1. 4F3/2
  2. 4F9/2
  3. 4D7/2
  4. 4D1/2

Answer (Detailed Solution Below)

Option 1 : 4F3/2

Atomic & Molecular Physics Question 9 Detailed Solution

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CONCEPT:

Spectral term: It is a symbol that represents the state of an atom.

An atomic state with spin 's' angular momentum 'l' and total momentum j canbe represented as

⇒ Spectral term = 2s+1LJ 

Here, L corresponds to a symbol. like l = 0 corresponds to 'S', 

l = 1 corresponds to 'P', l = 2 corresponds to 'D' and l = 3 corresponds to 'F' etc.

EXPLANATION:

Ground state the energy has to follow the rule of highest S Highest L and lowest J.

Highest S = ∑ms = ½ + ½ + ½ = (3/2)

Highest L = | ∑ml | = |2+1+0| = 3 (corresponding symbol 'F')

Lowest J = |L - S| = |3 - 3/2| = 3/2 

∴ Spectral term = 2s+1LJ = (+1F3/2 = 4F3/2

Hence the correct answer is option 1.

Let the separation of the frequencies of the first Stokes and the first anti-Stokes lines in the pure rotational Raman Spectrum of the H2 molecule be Δv(H2), while the corresponding quantity for D2 is Δv( D2). The ratio Δv(H2)/Δv(D2) is

  1. 0.6
  2. 1.2
  3. 1
  4. 2

Answer (Detailed Solution Below)

Option 4 : 2

Atomic & Molecular Physics Question 10 Detailed Solution

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Explanation:

  • The hydrogen molecule H2 consists of two protons, while the deuterium molecule D2 consists of two deuterons, with deuterium being twice as heavy as hydrogen.
  • The energies of the rotational Raman transitions are inversely proportional to the reduced mass of the molecule. As deuterium is twice the mass of hydrogen, the transitions in hydrogen should occur at twice the frequency as those in deuterium.
  • This would make the ratio 

In a spectrum resulting from Raman scattering, let IR denote the intensity of Rayleigh scattering and IS and lAS denote the most intense Stokes line and the most intense anti-Stokes line, respectively. The correct order of these intensities is

  1. IS > IR > IAS
  2. IR > IS > IAS
  3. IAS > IR > IS
  4. IR > IAS > IS

Answer (Detailed Solution Below)

Option 2 : IR > IS > IAS

Atomic & Molecular Physics Question 11 Detailed Solution

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CONCEPT:

Rayleigh and Raman scattering:

  • Raman spectroscopy is a spectroscopic technique based on Raman scattering.
  • When a substance interacts with a laser beam (or any light wave), almost all of the light produced is Rayleigh scattered light (elastic process).
  • However, a small percentage (about 0.000001%) of this light is Raman scattered (inelastic process). Raman scattering is a process, where incident light interacts with molecular vibrations in a sample.
  • The photons from the laser beam interact with the molecules and excite the electrons in them.
  • The excited electrons immediately fall down to the ground level.
  • As electrons lose energy and fall down to the ground state, they emit photons.

There are three different scenarios of how light can be re-emitted after energy had been absorbed by an electron: 

  • An electron falls down to the original ground state and there is no energy change, therefore light of the same wavelength is re-emitted. This is called Rayleigh scattering.
  • After being excited, an electron falls to a vibrational level, instead of the ground level. This means the molecule absorbed a certain amount of energy, which results in light being emitted in a longer wavelength than the incident light. This Raman scatter is called “Stokes”.
  • If an electron is excited from a vibrational level, it reaches a virtual level with higher energy. When the electron falls down to the ground level, the emitted photon has more energy compared to the incident photon, which results in shorter wavelength. This type of Raman scatter is called “Anti-Stokes”.

EXPLANATION:

As most of the produced are Rayleigh scattered light so rayleigh's scattered light is the most intense. The intensity of the Rayleigh line is thus always higher than the intensity of the stokes and Anti-stokes lines. Whereas the intensity of stokes is higher than anti-stokes.

So, Thus the correct formation is IR > IS > IAS

Hence the correct answer is option 2.

A negative muon, which has a mass nearly 200 times that of an electron, replaces an electron in a Li atom. The lowest ionization energy for the muonic Li atom is approximately

  1. the same as that of He
  2. the same as that of normal Li
  3. 200 times larger than that of normal Li
  4. the same as that of normal Be

Answer (Detailed Solution Below)

Option 3 : 200 times larger than that of normal Li

Atomic & Molecular Physics Question 12 Detailed Solution

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CONCEPT:

Ionization energy: The minimum energy to remove the electron from the last orbit of an atom. 

Mathematically, Ionization energy = E = 13.6  (for electron)

and E = 13.6  (for muon)

Here, μ = memμ/(me + mμ) = reduced mass

EXPLANATION:

Given, mass of muon = mμ = 200 me

and mass of  proton = mp = 1836 me

⇒ Ionization energy = E = 13.6 = A 

(Where, A = 13.6 )

For Normal Li-atom :- 

⇒ 

So, E ≈ A

For Muonic Li-atom :- 

So, E ≈ 197 A

Thus, The lowest ionization energy for the muonic Li atom is approximately 200 times larger than that of normal Li.

Hence the correct answer is option 3.

Atomic & Molecular Physics Question 13:

In the absorption spectrum of H-atom, the frequency of transition from the ground state to the first excited state is vH. The corresponding frequency for a bound state of a positively charged muon (μ+) and an electron is vμ. Using mμ = 10-28 kg, me = 10-30 kg and mp ≫ me, mμ, the value of (vμ - vH)/vH is

  1. 0.001
  2. -0.001
  3. -0.01
  4. 0.01

Answer (Detailed Solution Below)

Option 3 : -0.01

Atomic & Molecular Physics Question 13 Detailed Solution

Concept:

  • In a bound system of two particles of masses m₁ and m₂, the frequency of light absorbed or emitted during an electronic transition is inversely proportional to the reduced mass (μ) of the system, where 
  • For the Hydrogen atom (electron-proton system), because the mass of the proton is much larger than that of the electron (mₚ >> mₑ), the reduced mass is essentially the mass of the electron, i.e., 
  • For the muon-electron system (a muon and an electron), the reduced mass is . Since the mass of the muon is much larger than the mass of the electron, the denominator (mₑ + mₘ) is essentially the mass of the muon (mₘ).
  • Therefore, the reduced mass 

Explanation:

  • The frequency for the Hydrogen atom and the muon-electron system are inversely proportional to the respective reduced masses, i.e.,  and 
  • Therefore, the difference  is given by:
  • Substituting the above values,  
  • But we are interested in the reverse, i.e.,
  • So, the value of  .

Atomic & Molecular Physics Question 14:

The energy (in keV) and spin-parity values E (JP) of the low lying excited states of a nucleus of mass number A = 152, are 122(2+), 366(4+), 707 (6+), and 1125(8+). It may be inferred that these energy levels correspond to a

  1. rotational spectrum of a deformed nucleus
  2. rotational spectrum of a spherically symmetric nucleus
  3. vibrational spectrum of a deformed nucleus
  4. vibrational spectrum of a spherically symmetric nucleus

Answer (Detailed Solution Below)

Option 1 : rotational spectrum of a deformed nucleus

Atomic & Molecular Physics Question 14 Detailed Solution

Explanation:

  • The given energy levels and their respective spin-parity values indicate a 'rotational spectrum' of a 'deformed nucleus'.
  • The fact that the energy levels are increasing but not uniformly, and considering the spin and parity, strongly suggests a pattern consistent with a nuclear deformation leading to a rotational energy level scheme.
  • Furthermore, the parity remains positive across transitions, following the (2n)+ rule typical for even-even deformed nuclei. So together, these characteristics point to an even-even deformed nucleus exhibiting a rotational spectrum.

Atomic & Molecular Physics Question 15:

If the binding energies of the electron in the K and L shells of silver atom are 25.4 keV and 3.34 keV , respectively, then the kinetic energy of the Auger electron will be approximately

  1. 22 keV
  2. 9.3keV
  3. 10.5keV
  4. 18.7 keV

Answer (Detailed Solution Below)

Option 4 : 18.7 keV

Atomic & Molecular Physics Question 15 Detailed Solution

K E. . Auger electron is K.E = (EK - EL) - EL

= EK - 2EL

= (25.4 - 2 × 3.34) keV

= 18.7 keV

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