Nuclear Physics and RadioactivityMedium
Question
Two radioactive substances A and B of mass numbers 200 and 212 respectively, shows spontaneous $\alpha$-decay with same Q value of 1 MeV . The ratio of energies of $\alpha$-rays produced by A and $B$ is $\_\_\_\_$ .
Options
A.$\frac{2548}{2650}$
B.$\frac{2706}{2646}$
C.$\frac{2597}{2600}$
D.$\frac{2862}{2499}$
Solution
Sol. [Key Concept: The $\alpha$-particle kinetic energy is $K_{\alpha} = \frac{A - 4}{A}Q$; the ratio for $A = 200$ and $A = 212$ is $\frac{196/200}{208/212} = \frac{2597}{2600}$.]
$${k_{\alpha} = \left( \frac{A - 4}{\text{ }A} \right)Q }{\therefore\frac{k_{\alpha_{1}}}{k_{\alpha_{2}}} = \frac{196}{200} \times \frac{212}{208} = \frac{2597}{2600} }$$
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