The energy that should be added to an electron to reduce its de Broglie wavelength from 1 nm to 0.5 nm is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → thrice the initial energy
The de Broglie wavelength of an electron is \( \lambda=h/p \), so halving the wavelength doubles the momentum. Its kinetic energy is \( K=p^{2}/2m \), so the energy grows as the square of the momentum.
\( \begin{aligned} p' &= \frac{h}{\lambda/2} \\ &= 2p \end{aligned} \)
\( \begin{aligned} K' &= \frac{p'^{2}}{2m} \\ &= \frac{(2p)^{2}}{2m} \\ &= 4K \end{aligned} \)
The energy that must be added is:
\( \begin{aligned} \Delta K &= K'-K \\ &= 4K-K \\ &= 3K \end{aligned} \)
NCERT (2026-27):
Wave Nature of Matter – “… De Broglie proposed that the wave length λ associated with a particle of momentum p is given as”
\( \displaystyle \lambda=\frac{h}{p} \)\( \displaystyle =\frac{h}{mv} \qquad (11.5) \)