Question

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

Three times the initial energy

de Broglie wavelength $\lambda \propto \frac{1}{\sqrt{KE}}$
$⟹KE\propto \frac{1}{{\lambda }^2}$
$\frac{K{E}_2}{K{E}_1}={\left(\frac{{\lambda }_1}{{\lambda }_2}\right)}^2={\left(\frac{1}{0.5}\right)}^2$
We get $K{E}_2=4K{E}_1$
Thus added energy $\Delta KE=4K{E}_1-K{E}_1=3K{E}_1$