Question

Proton and alpha particle have same $K.E$. Which one has greater de-Broglie wavelength $?$

Answer 1

Since De-broglie Wavelength:-

${\lambda }_{pr}=\frac{h}{p}=\frac{h}{\sqrt{2m{E}_{pr}}}$

and

${\lambda }_{\alpha }=\frac{h}{p}=\frac{h}{\sqrt{2m{E}_{\alpha }}}$

Since, $E_{Proton}=E_{\alpha -particle}$

Hence $\lambda \propto \frac{1}{\sqrt{m}}$

$\therefore$ Particle with greater mass will have Lesser De-broglie Wavelength and vice-versa.

We have; $m_{\alpha }>m_{pr}$

$\Rightarrow {\lambda }_{pr}<{\lambda }_{\alpha }$