JEE Mains · Physics · STD 12 - Dual nature of radiation and matter
- JEE Mains 2024
- Hard
- MCQ
The de-Broglie wavelength of an electron is the same as that of a photon. If velocity of electron is \(25 \%\) of the velocity of light, then the ratio of \(K.E.\) of electron and \(K.E.\) of photon will be:
Pick an option to check your answer.
Answer & Solution
Correct Answer
(B) \(\frac{1}{8}\)
Step-by-step Solution
Detailed explanation
For photon
\(\mathrm{E}_{\mathrm{p}}=\frac{\mathrm{hc}}{\lambda_{\mathrm{p}}} \Rightarrow \lambda_{\mathrm{p}}=\frac{\mathrm{hc}}{\mathrm{E}_{\mathrm{p}}}\)
For electron
\( \lambda_{\mathrm{e}}=\frac{\mathrm{h}}{\mathrm{m}_{\mathrm{e}} \mathrm{v}_{\mathrm{e}}}=\frac{\mathrm{hv}_{\mathrm{e}}}{2 \mathrm{~K}_{\mathrm{e}}} \)
\( \text { Given } \mathrm{v}_{\mathrm{e}}=0.25 \mathrm{c} \)
\( \lambda_{\mathrm{e}}=\frac{\mathrm{h} \times 0.25 \mathrm{c}}{2 \mathrm{~K}_{\mathrm{e}}}=\frac{\mathrm{hc}}{8 \mathrm{~K}_{\mathrm{e}}}\)
Also \(\lambda_{\mathrm{p}}=\lambda_{\mathrm{e}}\)
\( \frac{\mathrm{hc}}{\mathrm{E}_{\mathrm{p}}}=\frac{\mathrm{hc}}{8 \mathrm{~K}_{\mathrm{e}}} \)
\( \frac{\mathrm{K}_{\mathrm{e}}}{\mathrm{E}_{\mathrm{p}}}=\frac{1}{8}\)
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