Question
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 ?
✓
Answer
$\lambda=\frac{{h}}{{mv}}=\frac{{h}}{\sqrt{2 {mE}}}, {mv}=\sqrt{2 {mE}}$
$\lambda \propto \frac{1}{\sqrt{{E}}}$ $\frac{\lambda_{2}}{\lambda_{1}}=\sqrt{\frac{{E}_{1}}{{E}_{2}}}=\frac{3}{4}, \lambda_{2}=0.75 \lambda_{1}$
$\frac{{E}_{1}}{{E}_{2}}=\left(\frac{3}{4}\right)^{2}$
${E}_{2}=\frac{16}{9} {E}_{1}=\frac{16}{9} {E} \quad\left({E}_{1}={E}\right)$
Extra energy given $=\frac{16}{9} {E}-{E}=\frac{7}{9} {E}$
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