The momentum carried by a photon of wavelength $\lambda$ is

Question

Accepted Answer

B. $\dfrac{h}{\lambda}$

1. A photon has energy $E = h
u$ and, since it is massless, $E = pc$ where $c$ is the speed of light.
2. Equate: $pc = h
u$, so $p = \dfrac{h
u}{c}$.
3. Using $c = 
u\lambda$ gives $p = \dfrac{h
u}{
u\lambda} = \dfrac{h}{\lambda}$.
4. The same expression $p = h/\lambda$ is also what de Broglie applied to MATTER.
Option A has wrong dimensions ($h\,\lambda$ would have units of action·length).
Option C is the reciprocal.
**Option D has an extra factor of $c\lambda$.**
_Source: syllabus Physics Part 2, Ch 11, §11.7 "Particle Nature of Light: The Photon" + §11.8 (de Broglie hypothesis), pp. 9–11._

Answer

$h\lambda$

Answer

$\dfrac{\lambda}{h}$

Answer

$h\lambda c$

The momentum carried by a photon of wavelength $\lambda$ is

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