Question

Photoelectric emission is observed from a metallic surface for frequencies $\mathrm{v}_{1}$ and $\mathrm{v}_{2}$ of the incident light rays $\left(\mathrm{v}_{1}>\mathrm{v}_{2}\right)$. If the maximum values of kinetic energy of
the photo-electrons emitted in the two cases are in ratio of 1: k, then the threshold frequency of the metallic surface is:
((A) $\dfrac{v_{1} k-v_{2}}{k-1}$
(B) $\dfrac{\mathrm{v}_{2} \mathrm{k}-\mathrm{v}_{1}}{\mathrm{k}-1}$
(C) $\dfrac{\mathrm{v}_{1} \mathrm{k}-\mathrm{v}_{2}}{1-\mathrm{k}}$
(D) $\dfrac{\mathrm{v}_{1}-\mathrm{v}_{2} \mathrm{k}}{\mathrm{k}-1}$

Answer 1

Hint: We know that the minimum frequency of light required to give the work function energy is called the threshold frequency. The work function can be calculated by: For photons with a greater frequency than this the extra energy given to electrons is in the form of kinetic energy. Study of the photoelectric effect led to important steps in understanding the quantum nature of light and electrons and influenced the formation of the concept of wave-particle duality. The photoelectric effect is also widely used to investigate electron energy levels in matter. The photoelectric effect is a phenomenon in physics. The effect is based on the idea that electromagnetic radiation is made of a series of particles called photons. When a photon hits an electron on a metal surface, the electron can be emitted. The emitted electrons are called photoelectrons.

Complete step by step answer
We know that the photoelectric emission is the process through which the free electrons are liberated from the surface of metals when it absorbs light. The photoelectric emission depends on the frequency of light and not on the intensity of light. This process is called photoelectric emission. Laws of photoelectric emission: (i) There is a definite cut off value of frequency below which electrons cannot be ejected by any substance. (ii) Number of emitted electrons are directly proportional to the intensity of light incident.
The photoelectric effect refers to what happens when electrons are emitted from a material that has absorbed electromagnetic radiation. Physicist Albert Einstein was the first to describe the effect fully, and received a Nobel Prize for his work. Photoelectric emission is the function of ejection of certain electrons. These electrons are ejected to light. It is ejected from a metal surface. The best material for such an emission is semiconductors of silicon.
Einstein's photoelectric equation can be derived using the fact that it is based on Planck's quantum theory. Equate the expression for the energy of the incident photon and maximum kinetic energy of the emitted photon and work function of the metal to get the required expression of Einstein's photoelectric.
We know that:
$\mathrm{hv}_{1}=\mathrm{hv}_{0}+\dfrac{1}{2} \mathrm{mu}_{1}^{2}.....(1)$
$\mathrm{hv}_{2}=\mathrm{hv}_{0}+\dfrac{1}{2} \mathrm{mu}_{2}^{2}......(2)$
$\dfrac{1}{2} \mathrm{mu}_{1}^{2}=\left(\dfrac{1}{\mathrm{k}}\right) \dfrac{1}{2} \mathrm{mu}_{2}^{2}.....(3)$
From equation (1)$\mathrm{hv}_{1}=\mathrm{hv}_{0}+\dfrac{1}{2 \mathrm{k}} \mathrm{mu}_{2}^{2}$
$\dfrac{1}{2} \mathrm{mu}_{2}^{2}=\mathrm{khv}_{1}-\mathrm{khv}_{0} \ldots…..(4)$
From equations (2) and (4), we get-
$\mathrm{hv}_{2}=\mathrm{hv}_{0}-\mathrm{khv}_{0}+\mathrm{khv}_{1}$
$\mathrm{v}_{0}(1-\mathrm{k})=\mathrm{v}_{2}-\mathrm{kv}_{1}$
$\mathrm{v}_{0}=\dfrac{\mathrm{kv}_{1}-\mathrm{v}_{2}}{(\mathrm{k}-1)}$

So, the correct answer is option B.

Note: We know that the four laws of photoelectric emission are given as:
LAW 1: For a given metal and frequency, the number of photoelectrons emitted is directly proportional to the intensity of the incident radiation.
LAW 2: For a given metal, there exists a certain frequency below which the photoelectric emission does not take place.
During thermionic emission electrons are emitted from the metal surface by providing heat energy, whereas, during photoelectric emission light energy is emitted when electrons are emitted from the surface of metal.