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  • A photon has the same wavelength as the de-Broglie wavelength of electron. Given that = speed of light

A photon has the same wavelength as the de-Broglie wavelength of electron. Given that \mathrm{c}= speed of light and \mathrm{v} = speed of electron. Which of the following relation is correct? [Here \mathrm{E}_{\mathrm{e}}= Kinetic energy of electron, \mathrm{E}_{\mathrm{ph}} = energy of photon, p_e= momentum of electron and p_{p h}= momentum of photon]

Option: 1

\frac{\mathrm{E}_{\mathrm{e}}}{\mathrm{E}_{\mathrm{ph}}}=\frac{2 \mathrm{c}}{\mathrm{v}}


Option: 2

\frac{E_e}{E_{p h}}=\frac{v}{2 c}


Option: 3

\frac{p_e}{p_{p h}}=\frac{2 c}{v}


Option: 4

\frac{\mathrm{p}_{\mathrm{e}}}{\mathrm{p}_{\mathrm{ph}}}=\frac{\mathrm{c}}{2 \mathrm{v}}


Answers (1)

best_answer

\begin{aligned} &\text{Given, }\\ & \lambda_{\mathrm{ph}}=\lambda_e \\ & \therefore \quad \frac{\mathrm{h}}{\mathrm{p}_{\mathrm{ph}}}=\frac{\mathrm{h}}{\mathrm{p}_{\mathrm{e}}} \\ & \therefore \quad \mathrm{p}_{\mathrm{ph}}=\mathrm{p}_{\mathrm{e}} \\ & \therefore \quad \frac{\mathrm{E}_{\mathrm{ph}}}{\mathrm{c}}=\frac{2 \mathrm{E}_{\mathrm{e}}}{\mathrm{v}} \\ & \therefore \quad \frac{\mathrm{E}_{\mathrm{e}}}{\mathrm{E}_{\mathrm{ph}}}=\frac{\mathrm{v}}{2 \mathrm{c}} \end{aligned}

Posted by

Gautam harsolia

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