The kinetic energy (in keV) of the alpha particle, when the nucleus at rest undergoes alpha decay, is

5319

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5422

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5707

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5818

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Solution

The correct option is A 5319
A
The KE(m KeV) of the alpha particle, when the nucleus at rest undergoes alpha decay is momentum conservation.
$\sqrt{2 m_{α} k_{1}} = \sqrt{2 m_{P_{b}} k_{2}}$
$\frac{k_{1}}{k_{2}} = \frac{m_{P_{b}}}{m_{α}} = \frac{205}{4}$
$k_{1} + k_{2} = Q$ value reaction
$k_{1} = \frac{205}{209} \times 5.4 M e V = 5319 k e V$

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The kinetic energy (in keV) of the alpha particle, when the nucleus at rest undergoes alpha decay, is

The correct option is A 5319AThe KE(m KeV) of the alpha particle, when the nucleus at rest undergoes alpha decay is momentum conservation. √2mαk1=√2mPbk2k1k2=mPbmα=2054k1+k2=Q value reaction k1=205209×5.4MeV=5319keV