Question

A nucleus at rest undergoes a decay emitting an $\alpha$particle of de-Broglie wavelength. $\lambda =5.75\times 10^{-15}$ m. If the mass of the daughter nucleus is 223.610 amu and that of the $\alpha -$ particle is 4.002 amu. Determine the total kinetic energy in the final state. Hence obtain the mass of the parent nucleus in amu. $1amu=931.470McV/c^2$

Answer 1

(a) Given mass of a-particle $m=4.002$ arau and mass of daughter nucleus,
$M=223.610amu,$
de-Broglie wavelength of $\alpha -$particle,
$\lambda =5.75\times 10^{-15}m$
So, momentum of a-particle would be
$p=\frac{h}{\lambda }=\frac{6.63\times 10^{-34}}{5.76\times 10^{-15}}kg-m/s$
or $p=1.151\times 10^{-19}kg-m/s$
From law of conservation of linear momentum, this
should also be equal to the linear momentum of the
daughter nucleus (in opposite direction). Let $K_1$ and $K_2$ be the kinetic energies of $\alpha -$ particle and
daughter nucleus. Then total kinetic energy in the final
state is
$K=K_1+K_2=\frac{p^2}{2m}+\frac{p^2}{2M}=\frac{p^2}{2}(\frac{1}{m}+\frac{1}{M})$
$K=\frac{p^2}{2}\left(\frac{M+m}{Mm}\right)$
$1amu=1.67\times 10^{-27}kg$
Substituting the values, we get
$K=10^{-12}J$
$K=\frac{10^{-12}}{1.6\times 10^{-13}}=6.25MeV$
or $K=6.25MeV.$
(b) Mass defect, $\Delta m=\frac{6.25}{931.470}=0.0067amu$
Therefore, mass of parent nucleus = mass of $\alpha ;-$ particle
$=(4.002+223.610+0.0067)$ amu
$=227.62$ amu
Hence, mass of parent nucleus is 227.62 amu.