GATE 2026 CY – Question 32

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 · Numerical answer

The HF microwave spectrum has successive lines separated by 41.11 cm−1. Its bond length in Å (two decimal places) is _____. Given masses H = 1.008 amu, F = 18.998 amu; 1 amu = $1.661\times10^{-27}$ kg; $h=6.626\times10^{-34}$ J s; $c=2.998\times10^8$ m/s.

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Correct answer: 0.92 to 0.94

Explanation

For a rigid rotor, the rotational energy levels are $E_J=BJ(J+1)$ (with $B$ in wavenumbers). The spectral lines for $J\to J+1$ are at $\tilde\nu=2B(J+1)$, so **successive lines are separated by $2B$**.

**Step 1: rotational constant.**
$$2B=41.11\text{ cm}^{-1}\;\Rightarrow\;B=20.555\text{ cm}^{-1}=2055.5\text{ m}^{-1}.$$

**Step 2: reduced mass.**
$$\mu=\frac{1.008\times18.998}{1.008+18.998}\text{ amu}=0.95717\text{ amu}=0.95717\times1.661\times10^{-27}=1.5898\times10^{-27}\text{ kg}.$$

**Step 3: bond length** from $B=\dfrac{h}{8\pi^2c\,\mu r^2}$:
$$r=\sqrt{\frac{h}{8\pi^2cB\mu}}=\sqrt{\frac{6.626\times10^{-34}}{8\pi^2(2.998\times10^8)(2055.5)(1.5898\times10^{-27})}}=9.255\times10^{-11}\text{ m}.$$

The bond length is **0.93 Å**.