In Mössbauer experiment, a source emitting at 14.4 KeV (3.48 × 10¹⁸ Hz) had to be moved towards absorber at 2.2 mms⁻¹ for resonance. The shift in the frequency between the source and the absorber is

(c) 25.5 MHz. Explanation: Δν = ν v/c = 3.48×10¹⁸ × 2.2×10⁻³/3×10⁸ = 2.55×10⁷ Hz = 25.5 MHz.

Q48 · CSIR-NET Chemistry, June 2013

Paper: CSIR-NET June 2013 · Subject: Inorganic Chemistry · Chapter: Inorganic Spectroscopy · Topic: Inorganic Spectroscopy – General · Marks: 2 · Difficulty: Medium

In Mössbauer experiment, a source emitting at $14.4 \mathrm{KeV}\left(3.48 \times 10^{18} \mathrm{Hz}\right)$ had to be moved towards absorber at $2.2 \mathrm{mms}^{-1}$ for resonance. The shift in the frequency between the source and the absorber is

(a)15.0 MHz

(b)20.0 MHz

(c)25.5 MHz

(d)30.0 MHz

Answer

Answer: C ✓ checked by 4AB · confidence high

Explanation

Δν = ν v/c = 3.48×10¹⁸ × 2.2×10⁻³/3×10⁸ = 2.55×10⁷ Hz = 25.5 MHz.

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