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  • An electron with speed and a photon with speed c have the same de-Broglie wavelength. If the kinetic energy and momentum of e

An electron with speed v and a photon with speed c have the same de-Broglie wavelength. If the kinetic energy and momentum of electron are \mathrm{E}_{\mathrm{e}}$ and $\mathrm{p}_{\mathrm{e}} and that of photon are \mathrm{E}_{\mathrm{ph}}$ and $\mathrm{p}_{\mathrm{ph}} respectively. Which of the following is correct?
 

Option: 1

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


Option: 2

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


Option: 3

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


Option: 4

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


Answers (1)

best_answer

\mathrm{\lambda_{e}= \lambda_{ph}}
\mathrm{\frac{h}{p_{e}}= \frac{h}{p_{ph}}}

\mathrm{p_{e}=p_{ph}\: \rightarrow (1)}

\mathrm{E_{e}=\frac{1}{2}m_{e}V^{2}}
\mathrm{E_{ph}=m_{ph}c^{2}}
\mathrm{\frac{E_{e}}{E_{ph}}= \frac{\left ( m_{e}V \right )V}{2\left ( m_{ph}c \right )c}}
\mathrm{\Rightarrow \frac{E_{e}}{E_{ph}}= \frac{V}{2c}}

The correct option is (2)

 

Posted by

seema garhwal

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