Question 705072

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 $\hbar$: $\omega = \Delta + \frac{\hbar\Delta^2}{2mc^2} = \Delta(1 + \frac{\hbar\Delta}{2mc^2})$.