JEE PYQ: Atomic Physics Question 34
Question 34 - 2020 (06 Sep Shift 1)
An electron, a doubly ionized helium ion ($He^{++}$) and a proton are having the same kinetic energy. The relation between their respective de-Broglie wavelengths $\lambda_e$, $\lambda_{He^{++}}$ and $\lambda_p$ is:
(1) $\lambda_e > \lambda_{He^{++}} > \lambda_p$
(2) $\lambda_e < \lambda_{He^{++}} = \lambda_p$
(3) $\lambda_e > \lambda_p > \lambda_{He^{++}}$
(4) $\lambda_e < \lambda_p < \lambda_{He^{++}}$
Show Answer
Answer: (3)
Solution
de-Broglie wavelength, $\lambda = \frac{h}{p} = \frac{h}{\sqrt{2m(KE)}}$. $\lambda \propto \frac{1}{\sqrt{m}}$. As $m_{He^{++}} > m_p > m_e$, therefore $\lambda_{He^{++}} < \lambda_p < \lambda_e$ or $\lambda_e > \lambda_p > \lambda_{He^{++}}$.
BETA
