Question

We can say that the energy of a photon of frequency $f$ is given by $E = hf$ , where $h$ is Planck's constant. The momentum of a photon is $P = \dfrac{h}{\lambda }$ where $\lambda$ is the wavelength of the photon. Then we may conclude that velocity of light is equal to:
(A) $\sqrt {\left( {\dfrac{E}{P}} \right)}$
(B) $\dfrac{E}{P}$
(C) $EP$
(D) ${\left( {\dfrac{E}{P}} \right)^2}$

Answer 1

Speed of any wave is $v = \lambda f$ . Where, $\lambda$ is the wavelength and $f$ is the frequency of the wave.
Find the ratio of the given energy and momentum formula. From the ratio speed of light can be obtained.

Complete Step By Step Answer:
It is given that the energy of a photon is $E = hf$ .
The momentum of the photon $P = \dfrac{h}{\lambda }$ .
Where, $h$ is the Planck’s constant
$f$ is the frequency of the photon
$\lambda$ is the wavelength of the photon
It is required to find the speed of light in terms of energy $E$ and momentum $P$ .
We know that speed of light $c = \lambda f$
From the energy equation, we have $h = \dfrac{E}{f}$ .
Substitute the above obtained value of $h$ in the momentum equation.
$P = \dfrac{E}{{\lambda f}}$
Substitute $\lambda f = c$
$\Rightarrow P = \dfrac{E}{c}$
Or $c = \dfrac{E}{P}$
Hence, the correct option is (B) $\dfrac{E}{P}$ .

Additional Information:
A photon is an elementary particle of light defined as an energy packet of light. Some properties of photons are discussed below;
-It behaves likes a particle as well as a wave simultaneously
-The rest mass is zero
-Travels with a constant speed of light
-The energy depends on its frequency. The higher the frequency, the more energy it has.

Note:
Light is an electromagnetic wave. The speed of light is constant in free space and is equal to $c = 3 \times {10^8}m{s^{ - 1}}$ . All the electromagnetic waves travel at a speed equal to the speed of light. The speed of light in a medium of refractive index $\mu$ is $v = \dfrac{c}{\mu }$ .