A nucleus at rest undergoes $α -$ decay emitting an $α -$ particle of mass $m_{α}$ and de-Broglie wavelength $λ$ . If mass of daughter nucleus is $57 m_{α}$ , then the total kinetic energy produced in the decay is given by

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$\frac{57}{58} \frac{h^{2}}{λ^{2} m_{α}}$

$\frac{29}{57} \frac{h^{2}}{λ^{2} m_{α}}$

$\frac{87}{56} \frac{h^{2}}{λ^{2} m_{α}}$

$\frac{58}{57} \frac{h^{2}}{λ^{2} m_{α}}$

answer is B.

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Detailed Solution

${KE}_{α} = \frac{p^{2}}{2 m_{α}} = \frac{1}{2 m_{α}} \times (\frac{h^{2}}{λ^{2}}) and Q = \frac{A}{A - 4} \times ({KE}_{α}) = \frac{232}{228} \times (\frac{h^{2}}{2 λ^{2} m_{α}}) = \frac{29}{57} \frac{h^{2}}{λ^{2} m_{α}}$

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