Question 705072

SCQMEDIUM

An atom of mass $m$ , initially at rest, resonantly absorbs a photon. It makes a transition from the ground state to an excited state and also gets a momentum kick. If the difference between the energies of the ground state and the excited state is $\hbar\Delta$ , the angular frequency of the absorbed photon is closest to

(A)

$\Delta (1 + \frac{3}{2} \frac{\hbar\Delta}{mc^2})$

(B)

$\Delta (1 + \frac{1}{2} \frac{\hbar\Delta}{mc^2})$

(C)

$\Delta (1 + \frac{\hbar\Delta}{mc^2})$

(D)

$\Delta (1 + 2 \frac{\hbar\Delta}{mc^2})$

Detailed Solution

Conservation of energy: $\hbar\omega = \hbar\Delta + \frac{p^2}{2m}$ . Conservation of momentum: $p = \frac{\hbar\omega}{c}$ . Substituting $p$ : $\hbar\omega = \hbar\Delta + \frac{(\hbar\omega/c)^2}{2m}$ . Approximating $\omega \approx \Delta$ in the second term gives $\hbar\omega \approx \hbar\Delta + \frac{(\hbar\Delta)^2}{2mc^2}$ . Dividing by $\hbar$ : $\omega = \Delta + \frac{\hbar\Delta^2}{2mc^2} = \Delta(1 + \frac{\hbar\Delta}{2mc^2})$ .

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