Question

Calculate the momentum of a photon of energy $6 \times {10^{ - 19}}J$.

Answer 1

Hint: The given problem is from modern physics. As we know that the momentum of a photon is given by the ratio of energy of photon and speed of light. So here we use this relation to find out the momentum of the photon.

Complete step by step solution:
The momentum of the photon can be calculated with the help of photon energy because it can be represented in the ratio of total energy of photon to the speed of light.
Given, Energy of the given photon $E = 6 \times {10^{ - 19}}J$
The relation between momentum of photon and energy of photon can be derived with the help of de-Broglie wavelength equation as
Momentum of photon $P = \dfrac{E}{c}$
Where, $c = 3 \times {10^8}m/s$(speed of light)
After putting the value of energy of the photon in this relation, we will get the value of momentum of the photon.
$P = \dfrac{{6 \times {{10}^{ - 19}}}}{{3 \times {{10}^8}}} \\
P = 2 \times {10^{ - 27}}kg - m/\sec \\ $
So this is the value of momentum of the photon for the given value of energy of the photon.
The physicist Louis de Broglie suggested that the particles might have both wave properties and particle properties. de Broglie reasoned that matter also can show wave-particle duality, just like light, since light can behave as a wave and as a particle. And also reasoned that matter would follow the same equation for wavelength as light which is given by
$\lambda = \dfrac{h}{p}$
Where p is the linear momentum, as shown by Einstein.

Note: De Broglie derived the following relationships as
(1) $E=hν$ for a photon and $λν=c$ for an electromagnetic wave.
(2) $E=mc^2$, means $λ=h/mc$, which is equivalent to $λ=h/p$.
Here m is the relativistic mass, and noth the rest mass; since the rest mass of a photon is zero.