Question

A stationary radioactive nucleus of mass 210units disintegrates into an $\alpha$- particle of mass 4units and a residual nucleus of mass 206units. If the kinetic energy of the $\alpha$- particle is E, then the kinetic energy of the residual nucleus is

• A. $\left(\frac{2}{105}\right)E$
• B. $\left(\frac{2}{103}\right)E$
• C. $\left(\frac{103}{105}\right)E$
• D. $\left(\frac{103}{2}\right)E$

Answer 1

The correct option is B

$\left(\frac{2}{103}\right)E$

Step 1: Given data:

Mass of stationary radioactive nucleus, $m=210units$

Mass of $\alpha$- particle, $m_{\alpha }=4units$

Mass of residual nucleus of mass, $m_r=206units$ and

If the kinetic energy of the $\alpha$- particle, $K_{\alpha }=E$

Step 2: Finding the kinetic energy of residual nucleus $\left(K_r\right)$

The kinetic energy in terms of momentum is given by $K=\frac{p^2}{2m}$

As the radioactive nucleus is stationary, momentum remains constant. Therefore, $K\propto \frac{1}{m}$

Hence, the ratio of kinetic energies of the residual nucleus to $\alpha$- particle is $\frac{K_r}{K_{\alpha }}=\frac{m_{\alpha }}{m_r}$

$\Rightarrow K_r=\frac{m_{\alpha }}{m_r}\times K_{\alpha }$

$=\frac{4}{206}\times E=\left(\frac{2}{103}\right)E$

Hence, the kinetic energy of the residual nucleus is $\left(\frac{2}{103}\right)E$.

Thus B is the correct option.