Q. $\lambda_{e},\lambda_{p}$ and $\lambda_{\alpha}$ are the de Broglie wavelengths of electron, proton and $\alpha$ particle. If all are accelerated by same potential, then
A) $\lambda_{e} < \lambda_{p} < \lambda_{\alpha}$
B) $\lambda_{e} < \lambda_{p} > \lambda_{\alpha}$
C) $\lambda_{e} > \lambda_{p} < \lambda_{\alpha}$
D) $\lambda_{e} > \lambda_{p} > \lambda_{\alpha}$
Explanation
(d) de Broglie wavelength
$\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mE}} \Rightarrow \lambda \propto \frac{1}{\sqrt{Em}}$
Now kinetic energy $E$ gained by a charged particle under potential $V$ is $E = qV$ given $V$ is same for the given three paticles
$\therefore E_{e} = eV;E_{p} = eV$
$E_{\alpha} = 2eV \Rightarrow E_{e} = E_{p} < E_{\alpha}\ $and $m_{e} < m_{p} < m_{\alpha}$
$\Rightarrow \lambda = \frac{h}{\sqrt{2m_{e}E_{e}}} > \frac{h}{\sqrt{2m_{p}E_{p}}} > \frac{h}{\sqrt{2m_{\alpha}E_{\alpha}}}$
$\lambda_{e} > \lambda_{p} > \lambda_{\alpha}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.