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The momentum carried by a photon of wavelength $\lambda$ is
A$h\lambda$
B$\dfrac{h}{\lambda}$
C$\dfrac{\lambda}{h}$
D$h\lambda c$
Answer & Solution
Correct answer: B. $\dfrac{h}{\lambda}$
1. A photon has energy $E = h\nu$ and, since it is massless, $E = pc$ where $c$ is the speed of light.
2. Equate: $pc = h\nu$, so $p = \dfrac{h\nu}{c}$.
3. Using $c = \nu\lambda$ gives $p = \dfrac{h\nu}{\nu\lambda} = \dfrac{h}{\lambda}$.
4. The same expression $p = h/\lambda$ is also what de Broglie applied to MATTER (NCERT §11.8). 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: NCERT Class 12 Physics Part 2, Ch 11, §11.7 "Particle Nature of Light: The Photon" + §11.8 (de Broglie hypothesis), pp. 9–11._
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