A stationary nucleus of mass 24amu emits a gamma photon. The energy of the emitted photon is 7 MeV. The recoil energy of nucleus (in MeV) is?

>*I know that I must use conservation of momentum. But I am not able to do that.* **Anyone please help**

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First note that a 7MeV emmission will have a negligible effect effect on the mass of the nucleus so we can assume the mass of the nucleus is constant at 24amu.

Now note the momentum of the emitted gamma radiation is given by $p_g = \frac{E_\gamma}{c} $ where $E_\gamma$ is the energy of the emitted gamma radiation (in this case 7 MeV). Therefore,

$p_g^2= \frac{E_\gamma^2}{c^2}$

Also note the final momentum of the nucleus is given by $p_n = mv$ and therefore $p_n^2 = (mv)^2 = 2mE$ where $E$ is the kinetic energy of the nucleus and $m$ is the mass of the nucleus. Therefore,

$p_n^2 = 2mE_n$

By conservation of momentum, because initial momentum is 0 (nucleus is initially stationary), we must have,

$p_g = p_n$

And therefore,

$p_g^2 = p_n^2$

Therefore,

$\frac{E_\gamma^2}{c^2} = 2mE_n$

The recoil energy is therefore given by,

$E_n = \frac{E_\gamma^2}{2mc^2}$

If the energy is given in units of MeV and the mass in amu, it can be shown that,

$E_n = \frac{E_\gamma^2}{1862m} = \frac{7^2}{1862\times 24} = 1.1\times10^-3 MeV = 1.1KeV$

Thanks for your answer.

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