Q: If h is Planck’s constant is SI system, the momentum of a photon of wavelength 0.01 A° is:
(A) 10–2 h
(B) h
(C) 102
(D) 1012h
Solution : $\large \lambda = \frac{h}{p}$
$\large p = \frac{h}{\lambda}$
$\large p = \frac{h}{0.01 \times 10^{-10}}$
p = 1012h