Question

A stationary $P{b}^{200}$ nucleus emits an alpha-particle with kinetic energy $K_{\alpha }$. The fraction of recoil energy of the daughter nucleus to the total energy liberated is :

• A. $\frac{1}{196}$
• B. $\frac{4}{196}$
• C. $\frac{1}{20}$
• D. $\frac{1}{50}$

Answer 1

The correct option is D $\frac{1}{50}$

$P{b}^{200}\to x^{196}{+}_{2}H{e}^4$
By conserving momentum

$0=4V_1+196V_2$

$V_1=-\frac{196}{4}{V}_2$

$V_1=-49V_2$
Total kinetic energy of emitted nuclei

$K_T=\frac{1}{2}\times 4\times V_1^2+\frac{1}{2}\times 196V_2^2$

Kinetic energy of daughter nuclei $x^{196}$

$K_D=\frac{1}{2}\times 196V_2^2$

$\frac{K_D}{K_T}=\frac{196V_2^2}{4V_1^2+196V_2^2}=\frac{49V_2^2}{V_1^2+49V_2^2}$

$=\frac{49V_2^2}{V_2^2[\left(\frac{V_1^2}{V_2^2}\right)+49]}=\frac{49}{(49{)}^2+49}$

$=\frac{1}{49+1}=\frac{1}{50}$