Modern Physics Ques 183
- A nucleus with mass number 220 initially at rest emits an $\alpha$-particle. If the $Q$ value of the reaction is $5.5 MeV$, calculate the kinetic energy of the $\alpha$-particle
$(2003,2 M)$
(a) $4.4 MeV$
(b) $5.4 MeV$
(c) $5.6 MeV$
(d) $6.5 MeV$
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Answer:
Correct Answer: 183.(b)
Solution:
Formula:
- Given that $K _1+K _2=5.5 MeV$
From conservation of linear momentum,
$$ \begin{aligned} & p _1=p _2 & & \\ \text { or } & \sqrt{2 K _1(216 m)} & =\sqrt{2 K _2(4 m)} & \text { as } \quad p=\sqrt{2 K m} \\ \therefore & K _2 & =54 K _1 & \end{aligned} $$
Solving Eqs. (i) and (ii), we get
$K _2=KE$ of $\alpha$-particle $=5.4 MeV$.
BETA
