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 ?

• A. $\lambda_{e}=\lambda_{p}=\lambda_{\alpha}$
• B. $\lambda_{e} < \lambda_{p} < \lambda_{\alpha}$
• C. $\lambda_{e} > \lambda_{p} > \lambda_{\alpha}$
• D. $\lambda_{e}=\lambda_{p} > \lambda_{\alpha}$

Answer 1

Answer: (C)

$\lambda=\frac{h}{p}=\frac{h}{\sqrt{2 m E}} \propto \frac{1}{\sqrt{m}}$
$m_{a} > m_{p} > m_{e}$
$SO \lambda_{e} > \lambda_{p} > \lambda_{a}$