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A nucleus with mass number 220 initially at rest emits an $α -$ particle. If energy released in the reaction is 5.5 MeV. The K.E of the $α -$ particle is

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5.4 MeV

4.5 MeV

6.5 MeV

5.6 MeV

answer is A.

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Detailed Solution

$0 = \bar{P_{α}} + \bar{P_{D}}$
$\bar{P_{α}} = - \bar{P_{D}}$
$\frac{K . E_{α}}{K . E_{D}} = \frac{m_{D}}{m_{α}} = \frac{216}{4} = \frac{54}{1}$
$K . E_{α} = \frac{54}{55} (T . E) = 5.4 M e V$

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