Q148 / 180 · KCET 2013 · Physics
The additional energy that should be given to an electron to reduce its de-Broglie wavelength from $1 \mathrm{~nm}$ to $0.5 \mathrm{~nm}$ is
- A2 times the initial kinetic energy
- Correct answerB3 times the initial kinetic energy
- C$0.5$ times the initial kinetic energy
- D4 times the initial kinetic energy
Solution
The de-Broglie wavelength,
$\lambda=\frac{h}{\sqrt{2 m E_{k}}} \quad \therefore \lambda \propto \frac{1}{\sqrt{E_{k}}}$
( $\because$ h and $m$ remains constant)
Here, in given condition
$\begin{array}{ll}
\frac{\lambda_{1}}{\lambda_{2}}=\sqrt{\frac{E_{k_{2}}}{E_{k_{1}}}}, \quad \frac{\lambda_{1}}{\lambda_{2}}=\sqrt{\frac{4 E}{E}} \\
\lambda_{1}=2 \lambda_{2}, & \lambda_{2}=\frac{\lambda_{1}}{2}
\end{array}$
So, if $E_{k}$ is increased by four times, then $\lambda$ becomes half.
$\therefore$ Additional kinetic energy that should be supplied to the electron
$=4 E_{k}-E_{k}=3 E_{k}$
Hence, the correct answer is option: 'B'.