Q 12-11-135JEE MainJEE Main 2021 (26 Aug, Shift 2)Easy
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: (B) $\frac{7E}{9}$
$\lambda = \dfrac{h}{\sqrt{2mE}} \Rightarrow E \propto \dfrac{1}{\lambda^2}$.
$E' = \dfrac{E}{(0.75)^2} = \dfrac{16E}{9}$, so extra energy $= \dfrac{16E}{9} - E = \dfrac{7E}{9}$.
Solution by Sreeraj P, M.Sc Physics