Physics - Modern Physics Question with Solution | TestHub

The de-Broglie wavelength associated with a charged particle is given by $$\lambda =\frac{\mathrm{h}}{\mathrm{p}}$$ where, $\mathrm{h}=$ planck's constant and $\mathrm{p}=$ momentum $=\sqrt{2\mathrm{mKE}}$ (here, KE is the kinetic energy of the charged particle) $$\Rightarrow \lambda =\frac{\mathrm{h}}{\sqrt{2\mathrm{mKE}}}$$ For proton and $\alpha$-particle, the wavelengths are respectively given as, $$\begin{array}{r}{\lambda }_{\mathrm{p}}=\frac{\mathrm{h}}{\sqrt{2{\mathrm{m}}_{\mathrm{p}}{\mathrm{KE}}_{\mathrm{p}}}}\text{ and }{\lambda }_{\alpha }=\frac{\mathrm{h}}{\sqrt{2{\mathrm{m}}_{\alpha }{\mathrm{KE}}_{\alpha }}}\\ \therefore \frac{{\lambda }_{\mathrm{p}}}{{\lambda }_{\alpha }}=\frac{\sqrt{2{\mathrm{m}}_{\alpha }{\mathrm{KE}}_{\alpha }}}{\sqrt{2{\mathrm{m}}_{\mathrm{p}}{\mathrm{KE}}_{\mathrm{P}}}}\end{array}$$ Here, ${\mathrm{KE}}_{\alpha }={\mathrm{KE}}_{\mathrm{p}}$ and ${\mathrm{m}}_{\alpha }=4{\mathrm{m}}_{\mathrm{p}}$ Substituting these above mentioned relations in Eq. (i), we get $$\Rightarrow \frac{{\lambda }_{\mathrm{P}}}{{\lambda }_{\alpha }}=\sqrt{\frac{4{\mathrm{m}}_{\mathrm{P}}}{{\mathrm{m}}_{\mathrm{p}}}}=\frac{2}{1}\text{ or }{\lambda }_{\mathrm{P}}:{\lambda }_{\alpha }=2:1$$