Question

By photoelectric effect, Einstein proved
A. \[E = h\nu \]
B. \[K.E = \dfrac{1}{2}m{v^2}\]
C. \[E = m{c^2}\]
D. \[E = \dfrac{{Rh{c^2}}}{{{n^2}}}\]

Answer 1

Hint:The photoelectric effect is the physical process in which the electrons from the surface of metal get ejected by absorbing energy from the incident photon rays.

Formula used:
\[E = h\nu \],
Here E is the energy of the photon with frequency \[\nu \] and $h$ is the planck's constant.

Complete step by step solution:
Initially the photoelectric effect was assumed to be the confirmation of the wave nature of light. As per the wave theory of electromagnetic waves, the energy is distributed uniformly in the wave-front which comprises mutually perpendicular electric field and magnetic field.

Initially as per the wave theory when the intensity of the incident light was to be increased then the maximum kinetic energy of the ejected electron during photoelectric effect was supposed to be increased but experimentally this was not the result.

Then Einstein suggested that the energy carried by each particle of the light is directly proportional to the frequency of the light. He called the basic unit of electromagnetic wave energy packet a photon. Hence, the energy of a photon was proportional to the frequency of the electromagnetic wave of which the photon is a part of.
\[E \propto \nu \]
\[\Rightarrow E = h\nu \]
Here $h$ is the Plank’s constant to offer attributes to the Physicist Planck, who proposed the particle nature of the light.
Hence, Einstein proved \[E = h\nu \]

Therefore, the correct option is A.

Note: We already know that the relationship between frequency and wavelength is inverse. Therefore, threshold frequency is the lowest value of the photon's frequency that has enough energy to evict the electron from metal.