Question

Assertion: The energy (E) and momentum (p) of a photon are related by p=E/c.
Reason: The photon behaves like a particle.
A. Both A and R are true and R is the correct explanation of A
B. Both A and R are true but R is not correct explanation of A
C. A is true but R is false
D. A and R are false

Answer 1

Hint: Think about the nature of light. Also try to remember the energy equation valid for light and try to find out a momentum as the question. Next check the reason and assertion carefully and find out the write option matching with your decision.

Complete step by step solution:
To solve this question, we must know about some pages of the history of science. Let's start. Before 1900, scientists only understood electromagnetic radiation to be made up of waves. So, they also considered light as waves. After that world famous scientist Sir Albert Einstein did the photoelectric effect. Then we all came up in front of a new concept that is wave-particle duality of light which simply means that light is a wave as well as a particle. Because photons of light collide with mobile electrons of metals and due to collision electrons get some kinetic energy.

As we increase the frequency of light the energy of photons increases as well as kinetic energy increases. After gaining a particular amount of energy from the collision after a particular frequency electron comes up from the metal surface. Now here we think if light is a wave then it is not possible because collision between a particle and wave never happens. So here light acts as a particle. Now, it is clear why we say light as a wave as well as a particle.

So, taking this let’s compare two equations, $E = m{c^2}$ and $E = hv$. But here explaining again for more clarity that E = energy, m = mass, h = planck constant, v = frequency, c = velocity of light . $m{c^2} = hv$ by comparing we get. But break it more. We know that momentum of a particle $(p) = mc$ and $v = c/\lambda $.

Let’s put this equation on the equation. We get $pc = \dfrac{{hc}}{\lambda }$ now we can cancel out $c$ form both sides. $p = \dfrac{h}{\lambda }$ we get that. Now if we put this on $E = \dfrac{{hc}}{\lambda } = pc$ . By rearranging the equation $p = \dfrac{E}{c}$ which in our question. So, we can say that the assertion and the reason both are correct.

Hence option A is the correct answer.

Note: Light can act as a wave as well as a particle. So, both energy equations,$E = m{c^2}$ and $E = hv$ are valid for light. So, for light express his momentum as $p = \dfrac{E}{c}$ . there is another thing we have to note that de Broglie wavelength for light is $\lambda = \dfrac{h}{p}$.