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The momentum of a photon is $2 \times {10^{ - 16}}gm - cm/sec$. Its energy is
A. $0.61 \times {10^{ - 26}}erg$
B. $2.0 \times {10^{ - 26}}erg$
C. $6 \times {10^{ - 6}}erg$
D. $6 \times {10^{ - 8}}erg$


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Last updated date: 23rd Apr 2024
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Answer
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Hint:
Energy and momentum of a photon are closely related. Similarly, a photon's momentum and energy are related to its wavelength and frequency, respectively. The following formula can be used to determine a photon's energy : $E = pc$

Formula used:
 $E = pc$

Complete step by step solution:

In the question, we have given the momentum of a photon is $2 \times {10^{ - 16}}gm - cm/sec$,
In order to know that the elementary particle known as a photon has a zero rest mass and moves at the speed of light in a vacuum. With the help of the provided formula, Einstein explained the momentum $(p)$ of a photon.
The equation relates a photon's energy and momentum is as follow:
$E = pc$
Here, energy of the photon be $E$, momentum of the photon is$p = 2 \times {10^{ - 16}}gm - cm/sec$
The rate at which a light photon moves through space is referred to as the speed of light. It is denoted by a letter $c$and expressed in SI units of measurement $m/sec$. Though the speed of light or the value of $c$are constants. This physical constant has several applications in various areas of physics.
Then the velocity of the light in the vacuum is:
$c = 299792458\,m/sec \\$
 $\Rightarrow c \cong 3 \times {10^8}\,m/sec \\$
 $\Rightarrow c \cong 3 \times {10^{10}}\,cm/sec \\$
To determine the energy of the photon, substitute the given information in the above formula, then we obtain:
$E = (2 \times {10^{ - 16}}) \cdot (3 \times {10^{10}}) \\$
 $\Rightarrow E = 6 \times {10^{ - 6}}erg \\$
Thus, the correct option is: (C) $6 \times {10^{ - 6}}erg$



Note:
 It should be noted that the spin of light angular momentum is independent of its frequency because photons constantly move at the speed of light, the component measured along its direction of motion is the best approach to describe the spin. That’s why we need to apply an equation which relates a photon’s energy and momentum instead of its formula.