GATE 2025 CY – Question 61

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. · 2 marks · Numerical answer

If a molecule emitting a radiation of frequency $3.100\times10^9$ Hz approaches an observer with a relative speed of $5.000\times10^6\ \mathrm{m\,s^{-1}}$, then the observer detects a frequency of ____ $\times10^9$ Hz. (rounded off to three decimal places) [Given: Speed of light $c=3.000\times10^8\ \mathrm{m\,s^{-1}}$]

Practise this question in The GATE Grind →

Show answer and explanation

Correct answer: 3.136 to 3.168

Explanation

This is the **Doppler effect** for a source approaching the observer with speed $v$ comparable to the speed of light (here $v/c=5\times10^6/3\times10^8=0.01667$). Use the relativistic form for motion along the line of sight:
$$\nu_{obs}=\nu_0\sqrt{\frac{1+v/c}{1-v/c}} .$$

**Numbers.**
$$\frac vc=\frac{5.000\times10^6}{3.000\times10^8}=\frac1{60},\qquad\frac{1+1/60}{1-1/60}=\frac{61/60}{59/60}=\frac{61}{59}=1.03390 .$$
$$\sqrt{1.03390}=1.01681 .$$

$$\nu_{obs}=3.100\times10^9\times1.01681=3.152\times10^9\text{ Hz}.$$

The observer detects **3.152 × 10⁹ Hz**.