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A radioactive nucleus of mass $M$ emits a photon of frequency $v$ and the nucleus recoils. The recoil energy will be
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Verified Answer
The correct answer is:
$h^2 v^2 / 2 M c^2$
Momentum of a photon
$$
p=\frac{h v}{c}
$$
Hence, Recoil energy
$$
E=\frac{p^2}{2 M}
$$
$$
\begin{aligned}
& \therefore \quad E=\frac{\left(\frac{h v}{c}\right)^2}{2 M} \\
& \text { or } \quad E=\frac{h^2 v^2}{2 M c^2} \\
&
\end{aligned}
$$
$$
p=\frac{h v}{c}
$$
Hence, Recoil energy
$$
E=\frac{p^2}{2 M}
$$
$$
\begin{aligned}
& \therefore \quad E=\frac{\left(\frac{h v}{c}\right)^2}{2 M} \\
& \text { or } \quad E=\frac{h^2 v^2}{2 M c^2} \\
&
\end{aligned}
$$
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