Define the terms "stopping potential' and 'threshold frequency' in relation to the photoelectric effect. How does one determine these physical quantities using Einstein's equation?

#### Solution

**Stopping potential: **For a particular frequency of incident radiation, the minimum negative (retarding) potential V

_{0 }given to the anode plate for which the photocurrent stops or becomes zero is called the cut-off or stopping potential.

**Threshold frequency:**

There exists a certain minimum cut-off frequency ν_{0}, for which the stopping potential is zero and below ν_{0 }the electron emission is not possible.

This cut-off frequency is known as threshold frequency ν_{0}, which is different for different metal. In the photoelectric effect, an electron absorbs a quantum of energy (hν ) of radiation. If this quantum of energy absorbed by electron exceeds the minimum energy required to come out of the metal surface by electron, the kinetic energy of the emitted electron is

`"K" = "hv" - phi` ...(1)

Where `phi` is the minimum energy for electron to come out of the metal, and is different for different electrons in the metal. The maximum kinetic energy of photoelectrons is given by

`"K""max" = "hv" - phi0` ...(2)

Where, `phi0 - ` work function or least value of φ equation (2) is known as Einstein's photoelectric equation.

**Explanation of photoelectric effect with the help of Einstein's photoelectric equation**

**(i) **According to equation (2), K_{max }depends linearly on ν, and is independent of the intensity of radiation. This happens because, here, the photoelectric effect arises from the absorption of a single quantum of radiation by a single electron. The intensity of the radiation (that is proportional to the number of energy quanta per unit area per unit time) is irrelevant to this basic process.

**(ii) **Since K_{max} must be non-negative, equation (2) implies that photoelectric emission is possible only if h ν > `phi0`.

or v > v_{0}, where v_{0} = `"V"_0 = phi_0/"h"`

Thus, there exists a threshold frequency v_{0} `"V"_0 = phi_0/"h"` exists, below which photoelectric emission is not possible, and is independent of intensity.

**(iii)** As the intensity of radiation is proportional to the number of energy quanta per unit area per unit time. The greater the number of energy quanta available, the greater is the number of electrons absorbing the energy quanta, and therefore, the number of electrons coming out of the metal (for ν > ν_{0}) is more and so is photoelectric current.