Question

A nucleus at rest undergoes $\alpha -decay$ emitting an $\alpha -particle$ of de Broglie wavelength $\lambda =5.76\times {10}^{-15}$ 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 parent nucleus in amu. $(1amu=931.470MeV{c}^{-2})$

Answer 1

de Broglie wavelength of $\alpha -particle,{\lambda }_{\alpha }=5.76\times {10}^{-15}m$
de Broglie wavelength, $\lambda =\frac{h}{p}$
Therefore, momentum of $\alpha -particle$,
$p_{\alpha }=\frac{h}{\lambda }=\frac{6.63\times {10}^{-34}}{5.76\times {10}^{-15}}=1.15\times {10}^{-19}kgm{s}^{-1}$

If $p_d$ momentum of daughter nucleus, then from conservation of linear momentum,
$p_d=-p_{\alpha }=-1.15\times {10}^{-19}kgm{s}^{-1}$

Hence, total kinetic energy of final state,
$E=E_{\alpha }+E_d=\frac{p^2\alpha }{2m_{\alpha }}+\frac{p^2\alpha }{2m_{\alpha }}+\frac{p^{2}d}{2m_d}$
$=\frac{p^2\alpha }{2}(\frac{1}{m_{\alpha }}+\frac{1}{m_d})=\frac{p^2\alpha (m_d+m_{\alpha })}{2m_{\alpha }{m}_d}$
$1amu=931.470MeV/c^2=\frac{931.470\times 1.6\times {10}^{-13}}{(3\times {10}^8{)}^2}=kg$

Therefore, total kinetic energy of final state is:
$=\frac{(1.15\times {10}^{-19}{)}^2\times (223.610+4.002)amu}{2\times \left(4.002amu\right)\times \left(223.610amu\right)}$

$=\frac{(1.15\times {10}^{-19}{)}^2\times 227.612}{2\times 4.002\times 223.610\times 1.66\times {10}^{-27}}J$

$=\frac{(1.15{)}^2\times 227.612}{2\times 4.002\times 223.610\times 1.66}\times {10}^{-11}J$

$={10}^{-12}J=\frac{{10}^{-12}}{1.6\times {10}^{-13}}MeV=6.25MeV$

Mass of parent nucleus $=m_d+m_{\alpha }-\left(BE\right)$
$=(223.610+4.002-\frac{6.25}{931.47})amu$
$=(227.612-0.007)amu=227.605amu$