Question

Momentum of a photon having frequency \[1.5 \times {10^{13}}Hz\]
A. \[3.31 \times {10^{ - 29}}kgm/s\]
B. \[3.3 \times {10^{ - 34}}kgm/s\]
C. \[6.6 \times {10^{ - 34}}kgm/s\]
D. \[6.6 \times {10^{ - 30}}kgm/s\]

Answer 1

Hint: The photon is the qualitative unit of energy of the light wave, and the motion of the photon is characterized by the momentum of the photon. The momentum of the photon is inversely proportional to the wavelength of the light wave.

Formula used:
\[p = \dfrac{h}{\lambda }\]
where h is the Plank’s constant and p is the momentum of the photon with wavelength equals to \[\lambda \]
\[c = \nu \lambda \]
where c is the speed of light, \[\nu \] is the frequency of the photon and \[\lambda \] is the wavelength of the light wave.

Complete step by step solution:
The frequency of the light wave is the most characteristic property because it doesn’t change by changing the medium in which the light is travelling. The frequency of the photon is given as \[1.5 \times {10^{13}}Hz\].
\[\nu = 1.5 \times {10^{13}}Hz\]
We got the frequency of the photon and now we need to use the relation of the characteristic physical quantities for the electromagnetic wave to find the wavelength of the photon,
\[c = \nu \lambda \]
\[\Rightarrow \lambda = \dfrac{c}{\nu }\]

On putting the value of the speed of light and the frequency of the photon, we get
\[\lambda = \dfrac{{3 \times {{10}^8}}}{{1.5 \times {{10}^{13}}}}m\]
\[\Rightarrow \lambda = 2 \times {10^{ - 5}}m\]
Using the expression for the momentum of the photon, we find,
\[p = \dfrac{h}{\lambda }\]

Putting the values of Plank’s constant and the wavelength, we get
\[p = \dfrac{{\left( {6.626 \times {{10}^{ - 34}}} \right)}}{{\left( {2.00 \times {{10}^{ - 5}}} \right)}}kgm/s\]
\[\therefore p = 3.31 \times {10^{ - 29}}\,kgm/s\]
On rounding-off to the 2 decimal places, we get the momentum of the given photon as \[3.3 \times {10^{ - 29}}kgm/s\]

Therefore, the correct option is A.

Note: We must be careful about the units of the physical quantity while solving numerical problems. We need to convert all the given data into the standard unit form. The momentum makes the photon exert pressure at the surface of incidence. When there is a change in momentum of the photon, it exerts force on the surface of incidence.