The de-Broglie wavelength of a particle having kinetic energy $\mathrm{E}$ 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 ?
Correct Option: , 2
$\lambda=\frac{\mathrm{h}}{\mathrm{mv}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mE}}}, \mathrm{mv}=\sqrt{2 \mathrm{mE}}$
$\lambda \propto \frac{1}{\sqrt{E}}$
$\frac{\lambda_{2}}{\lambda_{1}}=\sqrt{\frac{\mathrm{E}_{1}}{\mathrm{E}_{2}}}=\frac{3}{4}, \lambda_{2}=0.75 \lambda_{1}$
$\frac{\mathrm{E}_{1}}{\mathrm{E}_{2}}=\left(\frac{3}{4}\right)^{2}$
$\mathrm{E}_{2}=\frac{16}{9} \mathrm{E}_{1}=\frac{16}{9} \mathrm{E}$
$\left(\mathrm{E}_{1}=\mathrm{E}\right)$
Extra energy given $=\frac{16}{9} \mathrm{E}-\mathrm{E}=\frac{7}{9} \mathrm{E}$
Ans. 2