Question

If the ratio of de-Broglle wavelength of a proton and an $\alpha -particle$ is $1:3$ , then

Answer 1

${\lambda }_{\alpha }=$ De-Broglle wave length of $\alpha$ particle

$\Rightarrow {\lambda }_{\rho }=$ De-Broglle wave length of proton

$\Rightarrow {\lambda }_{\alpha }=\frac{h}{\sqrt{2m_{\alpha }{k}_{\epsilon }}}$

$\Rightarrow {\lambda }_{\rho t}=\frac{h}{\sqrt{2m_{\rho }{k}_{\epsilon }}}$ $k_{\epsilon }=$ kinetic energy of proton

$\Rightarrow \frac{{\lambda }_{\rho }}{{\lambda }_{\alpha }}=\frac{h}{\sqrt{2m_{\rho }{k}_{\epsilon }}}\times \frac{\sqrt{2m_{\alpha }{k}_{\epsilon }}}{h}$

$=\sqrt{\frac{2m_{\alpha }{k}_{\epsilon }}{2m_{\rho }{k}_{\epsilon }}}$

$=\sqrt{\frac{m_{\alpha }}{m_{\rho }}}$

$=\sqrt{\frac{4m_{\rho }}{m_{\rho }}}$

$=\sqrt{\frac{4}{1}}$

$=\frac{2}{1}$

$=2:1$

Hence, the answer is $2:1.$