momentum vs wavelength

Question

If de Broglie wavelength halves, momentum:

RankGuru · Accepted Answer

From de Broglie's relation:
$p = \frac{h}{\lambda}$
- Momentum is **inversely proportional** to wavelength.
- If $\lambda$ becomes $\lambda/2$, the momentum becomes $h/(\lambda/2) = 2(h/\lambda) = 2p$.

Rank Guru

momentum vs wavelength

Select the correct option: