Q51 · CSIR NET Chemistry, December 2011
Paper: CSIR NET December 2011 · Subject: Inorganic Chemistry · Chapter: Inorganic Spectroscopy · Topic: Mossbauer · Marks: 2 · Difficulty: Medium
In $^{57}\mathrm{Fe}$ Mossbauer experiment, source of 14.4 keV (equivalent to $3.48 \times 10^{12} \mathrm{MHz}$) is moved towards absorber at a velocity of $2.2\ \mathrm{mm\,s^{-1}}$. The shift in frequency of the source for this sample is:
(c)20.2 MHz
(d)15.5 MHz
(a)35.5 MHz
(b)25.5 MHz
Answer
Answer: B ✓ checked by 4AB · confidence high
Explanation
Doppler shift Δν = ν v/c = 3.48×10¹² MHz × 2.2×10⁻³/3×10⁸ = 25.5 MHz.
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