Question

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

• A. 2 times the initial kinetic energy
• B. 3 times the initial kinetic energy
• C. 0.5 times the initial kinetic energy
• D. 4 times the initial kinetic energy

Answer 1

Answer: (B)

The de-Broglie wavelength,
$\lambda =\frac{h}{\sqrt{2m{E}_k}}\therefore \lambda \propto \frac{1}{\sqrt{E_k}}$
$(\because h$ and $m$ remains constant $)$
Here, in given condition
$\frac{{\lambda }_1}{{\lambda }_2}=\sqrt{\frac{E_{k_2}}{E_{k_1}}},\frac{{\lambda }_1}{{\lambda }_2}=\sqrt{\frac{4E}{E}}$
${\lambda }_1=2{\lambda }_2,{\lambda }_2=\frac{{\lambda }_1}{2}$
So, if $E_k$ is increased by four times, then $\lambda$ becomes half.
$\therefore$ Additional kinetic energy that should be supplied to the electron
$=4E_k-E_k=3E_k$