Question
The momentum of a photon of an electromagnetic radiation is $$3.3 \times {10^{ - 29}}kg{\text{ - }}m{s^{ - 1}}.$$ What is the frequency of the associated waves?
$$\left( {h = 6.6 \times {{10}^{ - 34}}J{\text{ - }}s,c = 3 \times {{10}^8}m{s^{ - 1}}} \right)$$
A.
$$1.5 \times {10^{13}}Hz$$
B.
$$7.5 \times {10^{12}}Hz$$
C.
$$6.0 \times {10^{13}}Hz$$
D.
$$3.0 \times {10^3}Hz$$
Answer :
$$1.5 \times {10^{13}}Hz$$
Solution :
The energy of a photon of a radiation of frequency $$\nu $$ and wavelength $$\lambda $$ is
$$E = h\nu = \frac{{hc}}{\lambda }\,.......\left( {\text{i}} \right)$$
If photon is considered to be a particle of mass $$m,$$ the energy associated with it, according to Einstein mass energy relation, is given by
$$E = m{c^2}\,.......\left( {{\text{ii}}} \right)$$
From Eqs. (i) and (ii),
$$\eqalign{
& h\nu = m{c^2} \cr
& {\text{or}}\,\,m = \frac{{h\nu }}{{{c^2}}} \cr
& \Rightarrow \frac{{h\nu }}{c} = mc \cr
& {\text{As}}\,\,mc = p\,\,\left( {p = {\text{momentum of photon}}} \right) \cr
& {\text{So,}}\,\,\nu = \frac{{cp}}{h} = \frac{{3 \times {{10}^8} \times 3.3 \times {{10}^{ - 29}}}}{{6.6 \times {{10}^{ - 34}}}} \cr
& = 1.5 \times {10^{13}}Hz \cr} $$