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JEE Mains · Physics · STD 12 - Dual nature of radiation and matter
- JEE Mains 2020
- Medium
- MCQ
A particle moving with kinetic energy \(E\) has de Broglie wavelength \(\lambda\). If energy \(\Delta \mathrm{E}\) is added to its energy, the wavelength become \(\frac {\lambda}{2} .\) Value of \(\Delta \mathrm{E},\) is
Pick an option to check your answer.
Answer & Solution
Correct Answer
(C) \(3E\)
Step-by-step Solution
Detailed explanation
de-Broglie wavelength \(=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mE}}}=\lambda\)
Also, \(\frac{\mathrm{h}}{\sqrt{2 \mathrm{m}(\mathrm{E}+\Delta \mathrm{E})}}=\frac{\lambda}{2}\)
\(\therefore \quad \frac{\mathrm{E}+\Delta \mathrm{E}}{\mathrm{E}}=4 \Rightarrow \Delta \mathrm{E}=3 \mathrm{E}\)
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