Question

The energy that should be added to an electron to reduce its de-Broglie wavelength from $1\, nm$ to $0.5\, nm$ is

• A. four-times the initial energy
• B. equal to the initial energy
• C. two-times the initial energy
• D. three-times the initial energy

Answer 1

Answer: (D)

We know that,
We have, $\lambda=\frac{h}{\sqrt{2 m k}}$
$\therefore $ Kinetic energy $(K E) \propto \frac{1}{\lambda^{2}}$
Hence, $\frac{K E_{2}}{K E_{1}}=\left(\frac{\lambda_{1}}{\lambda_{2}}\right)^{2}$
Here, $\lambda_{1} =1\, nm $
$\lambda_{2} =0.5 \,nm $
$\frac{K E_{2}}{K E_{1}} =\left(\frac{1}{0.5}\right)^{2} $
$K E_{2} =4 K E_{1}$
Hence, the kinetic energy is increases three times.
$\Delta K E=3 K E_{1}$