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JEE Mains · Physics · STD 12 - Dual nature of radiation and matter
  • JEE Mains 2021
  • Hard
  • MCQ

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 ?

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Answer & Solution

Correct Answer

(B) \(\frac{7}{9} {E}\)

Step-by-step Solution

Detailed explanation

\(\lambda=\frac{{h}}{{mv}}=\frac{{h}}{\sqrt{2 {mE}}}, {mv}=\sqrt{2 {mE}}\)

\(\lambda \propto \frac{1}{\sqrt{{E}}}\) \(\frac{\lambda_{2}}{\lambda_{1}}=\sqrt{\frac{{E}_{1}}{{E}_{2}}}=\frac{3}{4}, \lambda_{2}=0.75 \lambda_{1}\)

\(\frac{{E}_{1}}{{E}_{2}}=\left(\frac{3}{4}\right)^{2}\)

\({E}_{2}=\frac{16}{9} {E}_{1}=\frac{16}{9} {E} \quad\left({E}_{1}={E}\right)\)

Extra energy given \(=\frac{16}{9} {E}-{E}=\frac{7}{9} {E}\)

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