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An electron of mass ' $m$ ' and a photon have same energy $E$. The ratio of the de-Broglie wavelengths…
An electron of mass ' $m$ ' and a photon have same energy $E$. The ratio of the de-Broglie wavelengths associated with them is (c = velocity of light in air)
$\left[\frac{\mathrm{E}}{2 \mathrm{M}}\right]^{\frac{1}{2}}$ $\frac{1}{\mathrm{c}}\left[\frac{\mathrm{E}}{2 \mathrm{~m}}\right]^{\frac{1}{2}}$ $\mathrm{c}(2 \mathrm{mE})^{\frac{1}{2}}$ $\frac{1}{\mathrm{c}}\left[\frac{2 \mathrm{~m}}{\mathrm{E}}\right]^{\frac{1}{2}}$ Solution
De-broglie wavelength
For an electron, $\lambda_{\mathrm{e}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mE}}}$
For photon, $\mathrm{E}=\mathrm{pc} \Rightarrow \lambda_{\text {Photon }}=\frac{\mathrm{hc}}{\mathrm{E}}$
$\Rightarrow \frac{\lambda_{\mathrm{e}}}{\lambda_{\text {photon }}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mE}}} \times \frac{\mathrm{E}}{\mathrm{hc}}=\left(\frac{\mathrm{E}}{2 \mathrm{~m}}\right)^{1 / 2} \frac{1}{\mathrm{c}}$
Asked in: MHT CET 2022 (07 Aug Shift 1)
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