Question

A stationary $U^{238}$ nucleus undergoes $\alpha$- decay. If the kinetic energy of product nucleus is $E$, the total energy released in the process is-

• A. $E$
• B. $58.5E$
• C. $59.5E$
• D. $60.5E$

Answer 1

The correct option is C $59.5E$

mass of alpha particle $m_a=4m$, where $m$ is mass of a proton.

mass of Uranium nucleus is $m_u=238m$ after emission the residue will have mass $m_r=238-4=234m$

We know $KE=momentu{m}^2/2m$

so for residue nucleus, it should be like this $E=p_r^2/2m_r$

or momentum of residue nucleus is $p_r=\sqrt[2]{22m_{r}E}=\sqrt[2]{2468mE}$

But during emission the alpha particle should get same momentum in opposite direction to make net momentum as zero

before and after emission so it should also be $p_a=\sqrt[2]{2468mE}$

If kinetic energy of alpha particle is $E_a$ then $E_a=p_a^2/2m_a=\frac{468mE}{2\times 4m}=58.5E$

So net energy released in the emission is the sum of above two i.e. $59.5E$