Question

A particle, moving with a kinetic energy $E$, has de Broglie wavelength $\lambda$. If energy $\Delta E$ is added to its energy, the wavelength become $\frac{\lambda }{2}$. The value of $\Delta E$ is:

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

Answer 1

The correct option is C $3E$

As per question, when kinetic energy of the particle is $E$, wavelength is $\lambda$ and when kinetic energy becomes $E+\Delta E$, wavelength becomes $\frac{\lambda }{2}$.

Now, $\lambda =\frac{h}{\sqrt{2mE}}$

And, $\frac{\lambda }{2}=\frac{h}{\sqrt{2m(E+\Delta E)}}$

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

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

$\Rightarrow 4E=E+\Delta E$

$\therefore \Delta E=3E$

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