Nuclei $X$ decay into nuclei $Y$ by emitting $α$ particles. Energies of $α$ particle are found to be only $1 M e V$ & $1.4 M e V$ . Disregarding the recoil of nuclei $Y$ . The energy of $γ$ photon emitted will be:

0.8 M e V

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1.4 M e V

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1 M e V

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0.4 M e V

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Solution

The correct option is D $0.4 M e V$

$\begin{matrix} Given - * Nuclei X \to Nuclei Y & releasing α particles. * Energy of α particles \to IMev & 1.4 M e V . \end{matrix}$

$\begin{matrix} Since no external for u is applied on the Nuclei Hence, the momentum of the two α -particles must be same but here it is not so. So, some annount of energy also gets converted in γ -emission whose energy equals to the difference of the two α - partichs. \Rightarrow Energy = 1.4 M e V - 1 M e V = 0.4 M e V Hence, opt (d) is correct. \end{matrix}$

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