Q. The kinetic energy of electron and proton is $10^{- 32}J.$ Then the relation between their de-Broglie wavelength is
A) $\lambda_{p} < \lambda_{e}$
B) $\lambda_{p} > \lambda_{e}$
C) $\lambda_{p} = \lambda_{e}$
D) $\lambda_{p} = 2\lambda_{e}$
Explanation
(a) By using $\lambda = \frac{h}{\sqrt{2mE}}\ E = 10^{- 32}J =$ Constant for both particles. Hence $\lambda \propto \frac{1}{\sqrt{m}}$ Since $m_{p} > m_{e}$ so $\lambda_{p} < \lambda_{e}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.