JEE PYQ: Atomic Physics Question 38
Question 38 - 2020 (07 Jan Shift 2)
An electron (of mass $m$) and a photon have the same energy $E$ in the range of a few eV. The ratio of the de-Broglie wavelength associated with the electron and the wavelength of the photon is ($c$ = speed of light in vacuum)
(1) $\frac{1}{c}\left(\frac{2E}{m}\right)^{1/2}$
(2) $c(2mE)^{1/2}$
(3) $\frac{1}{c}\left(\frac{E}{2m}\right)^{1/2}$
(4) $\left(\frac{E}{2m}\right)^{1/2}$
Show Answer
Answer: (3)
Solution
De-Broglie wavelength of electron $\lambda_e = \frac{h}{\sqrt{2mE}}$ …(i) ($\because p = \sqrt{2mE}$). Energy of photon $E = \frac{hc}{\lambda_p}$. $\lambda_p = \frac{hc}{E}$ …(ii). Dividing (i) by (ii): $\frac{\lambda_e}{\lambda_p} = \frac{h}{\sqrt{2mE}} \cdot \frac{E}{hc} = \sqrt{\frac{E}{2m}} \cdot \frac{1}{c}$.
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