Q 12-11-140JEE MainJEE Main 2020 (9 Jan, Shift 1)Easy
A particle moving with kinetic energy $E$ has de Broglie wavelength $\lambda$. If energy $\Delta E$ is added to its energy, the wavelength becomes $\dfrac{\lambda}{2}$. Value of $\Delta E$, is:
Answer: (C) $3E$
$\lambda = \dfrac{h}{\sqrt{2mE}}$, so $\lambda \propto \dfrac{1}{\sqrt{E}}$.
Halving $\lambda$ needs four times the kinetic energy: $E + \Delta E = 4E$, so $\Delta E = 3E$.
Solution by Sreeraj P, M.Sc Physics