The energy that should be added to an electron to reduce its de-Broglie wavelength from \( 1\,\mathrm{nm} \) to \( 0.5\,\mathrm{nm} \) is:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → thrice the initial energy.
For an electron of kinetic energy \( K \), \( p = \sqrt{2mK} \), so the de Broglie wavelength is \( \lambda = h/\sqrt{2mK} \) and \( K \propto 1/\lambda^{2} \).
\( \displaystyle \frac{K_2}{K_1} = \left(\frac{\lambda_1}{\lambda_2}\right)^{2} = \left(\frac{1\,\mathrm{nm}}{0.5\,\mathrm{nm}}\right)^{2} = 4 \)
\( K_2 = 4K_1 \)
The energy to be added is the increase:
\( K_2 - K_1 = 4K_1 - K_1 = 3K_1 \)
So the energy added is thrice the initial energy.