Answer
The kinetic energy of alpha particle will be $(KE)_{\alpha}=1.6867\times10^{-13}J$
Work Step by Step
Given target nucleus diameter $=1.4\times10^{-14}m$
mass of alpha particle $m_{\alpha}=6.64\times10^{-27}kg$
kinetic energy of alpha particle $(KE)_{\alpha}=\frac{1}{2}m_{\alpha}v^2$
so $(KE)_{\alpha}=\frac{1}{2}\frac {(m_{\alpha}v)^2}{m_{\alpha}}$
here $v$ is speed of the alpha particle
momentum of alpha particle is given by $p_{\alpha}=m_{\alpha}v$
so $(KE)_{\alpha}=\frac{1}{2}\frac {(p_{\alpha})^2}{m_{\alpha}}$..........(1)
given that $\alpha $ particle should have de Broglie wavelength $\lambda_{\alpha}$ equal to diameter of target nucleus
so $\lambda_{\alpha}=1.4\times10^{-14}m$
from the definition of de Broglie wavelength $\lambda=\frac{h}{p}$
$\lambda_{\alpha}=\frac{h}{p_{\alpha}}$
or $p_{\alpha}=\frac{h}{\lambda_{\alpha}}$, putting this to equation (1)
we will get $(KE)_{\alpha}=\frac{1}{2}\frac {(\frac{h}{\lambda_{\alpha}})^2}{m_{\alpha}}$
$(KE)_{\alpha}=\frac{h^2}{2\times m_{\alpha}\times \lambda_{alpha}^2}$......(2)
putting values in equation (2)
$h=6.626\times10^{-34} J.s$
$m_{\alpha}=6.64\times10^{-27}kg$
$\lambda_{\alpha}=1.4\times10^{-14}m$
$(KE)_{\alpha}=\frac{(6.626\times10^{-34} J.s)^2}{2\times 6.64\times10^{-27}kg\times (1.4\times10^{-14}m)^2}$
$(KE)_{\alpha}=1.6867\times10^{-13}J$
