Question:
What should be the order of arrangement of de-Broglie wavelength of electron $\left(\lambda_{e}\right)$, an $\alpha$-particle $\left(\lambda_{\alpha}\right)$ and proton $\left(\lambda_{p}\right)$ given that all have the same kinetic energy ?
Correct Option: , 3
Solution:
$\lambda=\frac{\mathrm{h}}{\mathrm{p}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mE}}} \propto \frac{1}{\sqrt{\mathrm{m}}}$
$m_{\alpha}>m_{p}>m_{e}$
so $\lambda_{e}>\lambda_{p}>\lambda_{\alpha}$