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. $1\, amu \, = 931.470Mc V/c^2$

Answer 1

(a) Given mass of a-particle $m = 4.002$ arau and mass of daughter nucleus,
$M = 223.610 amu,$
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
$K = K_1 + K_2 = \frac {p^2}{2m} + \frac {p^2}{2M} = \frac {p^2}{2} \bigg(\frac {1}{m} + \frac{1}{M}\bigg)$
$K = \frac {p^2}{2}\bigg(\frac {M+m}{Mm} \bigg)$
$ 1 \, amu = 1.67 \times 10^{-27} kg$
Substituting the values, we get
$K = 10^{-12} J$
$K = \frac {10^{-12}}{1.6 \times 10^{-13}} = 6.25 MeV$
or $ \, \, \, \, \, \, \, \, K = 6.25 MeV.$
(b) Mass defect, $\Delta m = \frac {6.25}{931.470} = 0.0067 amu$
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.