Physics - Modern Physics Question with Solution | TestHub

KE of proton and alpha-particle shall be$1 \mathrm{eV}$and$2 \mathrm{eV}$respectively.

Now, momentum,$\text{p}=\sqrt{2\text{mKE}}$

∴ The momentum of proton is${\text{p}}_{\text{P}}=\sqrt{2{\text{m}}_p\times \text{eV}}$

∴ The momentum of α-particle whose mass

${\text{m}}_{\alpha }=4{\text{m}}_{\text{P}}$is

${\text{p}}_{\alpha }=\sqrt{8{\text{m}}_p\times \left(2\text{eV}\right)}=\sqrt{16{\text{m}}_p\text{eV}}$

∴ Now,$\lambda =\frac{\text{h}}{\text{p}}$[de-Broglie relation]

$\Rightarrow \frac{{\lambda }_{\text{p}}}{{\lambda }_{\alpha }}=\frac{{\text{p}}_{\alpha }}{{\text{p}}_p}=\frac{\sqrt{16{\text{m}}_p\text{eV}}}{\sqrt{2{\text{m}}_p\text{eV}}}=2\sqrt{2}\text{:}1$