Question:
A particle moving with kinetic energy $\mathrm{E}$ has de Broglie wavelength $\lambda$. If energy $\Delta \mathrm{E}$ is added to its energy, the
wavelength become $\frac{\lambda}{2}$. Value of $\Delta \mathrm{E}$, is:
Correct Option: , 3
Solution:
(3) As per question, when $\mathrm{KE}$ of particle $E$, wavelength $\lambda$ and when $K E$ becomes $E+\Delta E$ wavelength becomes $\lambda / 2$
Using, $\lambda=\frac{h}{\sqrt{2 m K E}}$
$\frac{\lambda}{2}=\frac{h}{\sqrt{2 m(K E+\Delta E)}}$
$\Rightarrow \frac{\lambda}{/ 2}=\sqrt{\frac{K E+\Delta E}{K E}}$
$\Rightarrow 4=\frac{K E+\Delta E}{K E}$
$\Rightarrow \quad 4 K E-K E=\Delta E$
$\therefore \quad \Delta E=3 \underline{K E}=3 \underline{E}$