Question

In the beta decay of Na${}^{24}$, the combined electron neutrino momentum has a magnitude equal to 4 $\frac{Mev}{c}$. What is the recoil energy of the daughter nucleus, given that its mass is $23.99u?$
Take $[c=3\times {10}^8\text{m/s},1u=1.66\times {10}^{-27}\text{kg}]$

• A. $0.72\times {10}^{-4}$ MeV
• B. $2.78\times {10}^{-4}$ MeV
• C. $1.25\times {10}^{-4}$ MeV
• D. $3.56\times {10}^{-4}$ MeV

Answer 1

The correct option is D $3.56\times {10}^{-4}$ MeV

Here, momentum of the electron and neutrino pair is
$p=4\frac{MeV}{c}=\frac{4\times 1.6\times {10}^{-13}}{3\times {10}^8}$
$p=2.13\times {10}^{-21}\frac{\text{kgm}}{s}$
Also mass of the daughter nucleus is
$m=23.99u=23.99\times 1.66\times {10}^{-27}\text{kg}$


As parent nucleus decays at rest, so according to conservation of linear momentum, the momentum of daughter nucleus and momentum of electron neutrino pair must be same
If $P^{\prime }$ $\Rightarrow$ momentum of daughter nucleus then,
$P^{\prime }=p$
Now recoil energy $E_r$ is
$E_r$ = $\frac{P^{\prime }2}{2m}=\frac{p^2}{2m}$
$\Rightarrow \frac{(2.13\times {10}^{-21}{)}^2}{2\times 23.99\times 1.66\times {10}^{-27}}\text{J}$
$\Rightarrow \frac{2.13\times 2.13\times {10}^{-15}}{2\times 23.99\times 1.66\times \times 1.66\times {10}^{-13}}\text{MeV}$

$\Rightarrow E=3.56\times {10}^{-4}\text{MeV}$