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GATE 2025 CY – Question 19

Physical Chemistry · Spectroscopy: Atomic spectroscopy; Russell-Saunders coupling; Term symbols and spectral details; origin of selection rules. Rotational, Vibrational, Electronic and Raman spectroscopy of diatomic and simple polyatomic molecules. Line broadening and line widths; simple properties of Gaussian and Lorentzian line shapes. Molecular spectroscopy: Absorbance, Beer- Lambert’s law, Einstein’s coefficient, Jablonski diagram. Relationship of transition moment integral with molar extinction coefficient and oscillator strength. Basic principles of Nuclear Magnetic Resonance: Gyromagnetic ratio; Chemical shift, nuclear coupling. · 1 mark · Multiple choice

Rigid rotor wavefunctions are given by $Y_{l,m}(\theta,\phi)$. The wavefunctions $Y_{1,0}(\theta,\phi)$ and $Y_{2,0}(\theta,\phi)$ are given below $$Y_{1,0}=\sqrt{\frac3{4\pi}}\cos\theta,\quad Y_{2,0}=\sqrt{\frac5{16\pi}}(3\cos^2\theta-1).$$ For a non-polar diatomic molecule, the value of transition dipole moment integral for transition between $Y_{1,0}$ and $Y_{2,0}$ is equal to

  1. $1/\sqrt{2\pi}$
  2. 0
  3. 2
  4. $1/\sqrt{4\pi}$

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Correct answer: (B) 0

Explanation

The transition dipole moment for a rotational transition is
$$\mu_{fi}=\int Y_f^*\,\hat\mu\,Y_i\,d\tau,$$
where $\hat\mu$ is the electric dipole moment operator.

**For a non-polar diatomic molecule** (such as $N_2$ or $O_2$) the permanent dipole moment is **zero**, and it does not change as the molecule rotates. The operator $\hat\mu=\mu_0\cos\theta$ is proportional to the permanent dipole $\mu_0$, so
$$\mu_{fi}\propto\mu_0=0 .$$

The integral is therefore **zero** whatever the angular overlap of $Y_{1,0}$ and $Y_{2,0}$ is. This is why homonuclear diatomics have no pure rotational (microwave) spectrum.

Answer **0** (B).