Question

Plutonium undergoes $\alpha$-decay to form uranium:
$23994\mathrm{Pu}\to 23592U+42\mathrm{He}$
The atomic masses of the elements are as follows:

Calculate the kinetic energy imparted to the daughter nuclei due to the mass defect.

• A. $6.74\text{MeV}$
• B. $10.44\text{MeV}$
• C. $5.22\text{MeV}$
• D. $2.61\text{MeV}$

Answer 1

The correct option is C $5.22\text{MeV}$

Given:
$23994\mathrm{Pu}\to 23592U+42\mathrm{He}$

• ${\text{m}}_{\text{Pu}}=239.0522\text{a.m.u.}$
• ${\text{m}}_{\text{U}}=235.044\text{a.m.u.}$
• ${\text{m}}_{\text{He}}=4.0026\text{a.m.u.}$

Change in the mass, $\Delta \text{m}={\text{m}}_{\text{Pu}}-({\text{m}}_{\text{U}}+{\text{m}}_{\text{He}})$
$⟹239.0522-(235.044+4.0026)$
$⟹0.0056\text{a.m.u.}$
$=5.6\times {10}^{-3}\text{a.m.u.}$
This apparent loss in the mass will manifest itself as the kinetic energy of the daughter nuclei, according to Einstein's mass-energy equivalence, $\text{E =}\Delta {\text{mc}}^2$ ($\text{c}$ is the speed of light $=3\times {10}^8{\text{m s}}^{-1}$)
$1\text{a.m.u.}$ of mass defect will result in an energy of about $932\text{MeV}$.
$⟹5.6\times {10}^{-3}\text{a.m.u.}$ would result in about $5.6\times {10}^{-3}\times 932$
$=5.2192\text{MeV}\approx 5.22\text{MeV}$ of energy.