Question

An $\alpha$-particle, a proton and an electron have the same kinetic energy. Which one of the following is correct in case of their de-Broglie wavelength:

• A. $\lambda_\alpha>\lambda_p>\lambda_e$
• B. $\lambda_\alpha<\lambda_p<\lambda_e$
• C. $\lambda_\alpha=\lambda_p=\lambda_e$
• D. $\lambda_\alpha>\lambda_p<\lambda_e$

Answer 1

Answer: (B)

$ \lambda_{ D }=\frac{ h }{ p }=\frac{ h }{\sqrt{2 mK }} $
$\therefore \lambda \propto \frac{1}{\sqrt{ m }} $
$ \because m _\alpha> m _{ p }> m _{ e }$
$ \lambda_{ c }>\lambda_{ p }>\lambda_\alpha$