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JEE Mains · Physics · STD 12 - 11. Dual nature of radiation and matter
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
- A \(2E\)
- B \(E\)
- C \(3E\)
- D \(4E\)
Answer & Solution
Correct Answer
(C) \(3E\)
Step-by-step Solution
Detailed explanation
de-Broglie wavelength \(=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mE}}}=\lambda\) Also, \(\frac{\mathrm{h}}{\sqrt{2 \mathrm{m}(\mathrm{E}+\Delta \mathrm{E})}}=\frac{\lambda}{2}\)…
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