Question

The momentum of a photon is $3.3 \times {10^{ - 29}}kgm/s$ . Its frequency will be
A. $3 \times {10^3}Hz$
B. $6 \times {10^3}Hz$
C. $7.5 \times {10^{12}}Hz$
D. $1.5 \times {10^{13}}Hz$

Answer 1

Hint: Photon is an elementary particle and quanta of light. Photons carry electromagnetic radiation; they are continuously in motion. They move with the speed of light in a vacuum. Photons are electrically neutral and massless particles.

Complete answer:
The momentum of the photon is given $p = 3.3 \times {10^{ - 29}}kg - m/s$ and we have to calculate its frequency.
We have the following De ‘Broglie relation between momentum and the wavelength of a particle.
$\lambda = \dfrac{h}{p}$ ……………….. (1)
Here, $\lambda $ is the wavelength of a particle, h is Planck’s constant and $p$ is the momentum of the particle.
We also have a relation between speed, wavelength, and frequency.
$c = \lambda \nu $ ……………………(2)
Here, $c$ is the speed of light, $\lambda $ is the wavelength, and $\nu $ is the frequency.
Putting the value of $\lambda $, from equation (2), in equation (1).
$\dfrac{c}{\nu } = \dfrac{h}{p}$
On rearranging the above expression, we get the following formula for frequency.
$\nu = \dfrac{{cp}}{h}$
Let us substitute the values in the above equation.
$\nu = \dfrac{{3 \times {{10}^8} \times 3.3 \times {{10}^{ - 29}}}}{{6.6 \times {{10}^{ - 34}}}} = \dfrac{{9.9 \times {{10}^{13}}}}{{6.6}} = 1.5 \times {10^{13}}Hz$
Therefore, the frequency of the photon is $1.5 \times {10^{13}}Hz$.

Hence, the correct option is (D) $1.5 \times {10^{13}}Hz$.

Note:
De Broglie explained that with every matter particle a wave is associated and the wavelength is inversely proportional to the momentum of the particle. This means the wavelength of high-radiations is very low.
The energy of radiation is measured by the energy of a photon, which is the product of Planck’s constant and frequency.