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Q.

A nucleus of mass number 238 at rest emits an alpha particle with kinetic energy 5.4MeV. The Q value of reaction is

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a

5.58 MeV

b

5.49 MeV

c

5.67 MeV

d

5.50 MeV

answer is A.

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Detailed Solution

$\begin{array}{l} ∴ Q = K_{α} + K_{p} \\ = \frac{238}{234} \times K_{α} \\ = \frac{238}{234} \times 5 .4 MeV \end{array}$

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