Question

A particle moving with kinetic energy $E$, has de-Broglie wavelength $\lambda$. If energy $\Delta E$ is added to it, the wavelength become $\lambda /2$. The value of $\Delta E$ is,

• A. $2E$
• B. $E$
• C. $3E$
• D. $4E$

Answer 1

The correct option is C $3E$

Given that $(K.E{)}_1=E$

we know that, $\lambda =\frac{h}{\sqrt{2mE}}.....\left(1\right)$
(where $\lambda$ is de-Broglie wavelength)

When $K.E$ changes to $(E+\Delta E)$, the de-Broglie wavelength is $\frac{\lambda }{2}$

$\Rightarrow \frac{h}{\sqrt{2m(E+\Delta E)}}=\frac{\lambda }{2}....\left(2\right)$

Dividing $\left(1\right)$ by $\left(2\right)$, we get,

$\frac{\sqrt{E+\Delta E}}{E}=2$

$\frac{E+\Delta E}{E}=4\Rightarrow \Delta E=3E$

Hence, $\left(C\right)$ is the correct answer.