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JEE Mains · Physics · STD 12 - 11. Dual nature of radiation and matter
The de-Broglie wavelength of a particle having kinetic energy \(E\) is \(\lambda\). How much extra energy must be given to this particle so that the de-Broglie wavelength reduces to \(75 \,\%\) of the initial value ?
- A \(\frac{1}{9} {E}\)
- B \(\frac{7}{9} {E}\)
- C \({E}\)
- D \(\frac{16}{9} {E}\)
Answer & Solution
Correct Answer
(B) \(\frac{7}{9} {E}\)
Step-by-step Solution
Detailed explanation
\(\lambda=\frac{{h}}{{mv}}=\frac{{h}}{\sqrt{2 {mE}}}, {mv}=\sqrt{2 {mE}}\) \(\lambda \propto \frac{1}{\sqrt{{E}}}\) \(\frac{\lambda_{2}}{\lambda_{1}}=\sqrt{\frac{{E}_{1}}{{E}_{2}}}=\frac{3}{4}, \lambda_{2}=0.75 \lambda_{1}\)…
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