The de-Broglie wavelength of a particle having kinetic energy $E$ is $\lambda$. How much extra energ

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 ?\

Professor PC Thomas Classes · Answer

(b) : According to de-Broglie hypothesis 
$ \begin{array}{r} \lambda_{1}=\frac{h}{\sqrt{m v}}=\frac{h}{\sqrt{2 m E}} \ 	herefore m v=\sqrt{2 m E} ; \lambda \propto \frac{1}{\sqrt{E}} \end{array} $
 Let the new wavelength is $\lambda_{2}$ and energy is $E_{2}$.
 $ \begin{aligned} & \frac{\lambda_{2}}{\lambda_{1}}=\sqrt{\frac{E_{1}}{E_{2}}}=\frac{3}{4} \Rightarrow \lambda_{2}=0.75 \lambda_{1} \ 	herefore & \frac{E_{1}}{E_{2}}=\left(\frac{75}{100}ight)^{2}=\left(\frac{3}{4}ight)^{2} \ \Rightarrow & E_{2}=\frac{16}{9}, E_{1}=\frac{16}{9} E \end{aligned} $
 So, extra energy $=\frac{16}{9} E-E=\frac{7}{9} E$\