Q. The de-Broglie wavelength of an electron, $\alpha$- particle and proton all having the same kinetic energy is respectively given as $\lambda_{e},\ \lambda_{\alpha}\ \ and\ \lambda_{p}.$ Then which of the following is not true?
A) $\lambda_{e} < \lambda_{p}$
B) $\lambda_{p} > \lambda_{\alpha}$
C) $\lambda_{e} > \lambda_{\alpha}$
D) $\lambda_{\alpha} < \lambda_{p} < \lambda_{e}$
Explanation
(a) de-Broglie wavelength
$\lambda = \frac{h}{mv}\ \ \ or\ \ \ \ \ \lambda \propto \frac{1}{m}$
$\therefore\ \ \ \ \lambda_{e} \propto \frac{1}{m},\ \ \ \lambda_{\alpha}\ \propto \frac{1}{m_{\alpha}}\ \ and\ \lambda_{p} \propto \frac{1}{m_{p}}\ \ \ \ \ $
As we know that ${\ m}_{e} < m_{p} < m_{\alpha}$
So, $\lambda_{e}\ > \lambda_{p} > \lambda_{\alpha}$
Or $\lambda_{e} > \lambda_{\alpha}\ \ \ \ \ $or $\lambda_{p} > \lambda_{\alpha}\ \ \ \ \ $or $\lambda_{e} > \lambda_{p}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.