Question

A nucleus of mass number 220 decays by $\alpha$ - decay. The energy released in the reaction is 5eV. The kinetic energy of an $\alpha$ -particle is

• A. $\frac{1}{54}MeV$
• B. $\frac{27}{11}MeV$
• C. $\frac{54}{11}MeV$
• D. $\frac{55}{54}MeV$

Answer 1

The correct option is C $\frac{54}{11}MeV$

$K_{\alpha }=\frac{A-4}{A}Q$
$=\frac{220-4}{220}\times 5MeV=\frac{54}{11}MeV$.