Modern Physics Ques 191
- The kinetic energy (in $keV$ ) of the alpha particle, when the nucleus ${ } _{84}^{210} Po$ at rest undergoes alpha decay, is
(a) $5319$
(b) $5422$
(c) $5707$
(d) $5818$
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Answer:
Correct Answer: 191.(a)
Solution:
- ${ } _{84} Po^{210} \rightarrow{ } _2 He^{4}+{ } _{82} Pb^{206}$
$ \begin{aligned} \text { Mass defect } \Delta m & =\left(m _{Po}-M _{He}-m _{Pb}\right)=0.005818 u \\ \therefore \quad Q & =(\Delta m)(931.48) MeV=5.4193 MeV \\ & =5419 keV \end{aligned} $
From conservation of linear momentum,
$ \begin{aligned} p _{Pb} & =p _{\alpha} \\ \therefore \quad \sqrt{2 m _{Pb} k _{Pb}} & =\sqrt{2 m _{\alpha} k _{\alpha}} \text { or } \frac{k _{\alpha}}{k _{Pb}}=\frac{m _{Pb}}{m _{\alpha}}=\frac{206}{4} \\ \therefore \quad k _{\alpha} & =(\frac{206}{206+4})\left(k _{\text {total }}\right) \\ & =(\frac{206}{210})(5419)=5316 keV \end{aligned} $
BETA
