Class 12 Physics Dual Nature Radiation Matter | Einsteins Photoelectric Equation |

- Since wave theory could not explain the photoelectric effect, Einstein proposed a particle theory of light for the first time
- He said that radiations are made up of specific and discrete packets of energy called as
__quanta__of radiation energy. Each energy quantum has a value equal to, where__hv__*h = Planck’s constant*, and*v = frequency of incident light* - These specific packets of quanta of energy are known as
__photons__ - When a light of frequency(
*v*) (having energy*hv*) is incident on a metal surface of work function(*Φ*), 3 cases could be possible_{o} __Case-1__-When (*hv < Φ*), i.e., energy of photon is less than the work function of metal , no photoelectric emission occurs_{o}__Case-2__- When (*hv = Φ*), i.e., energy of photon is exactly same as the work function of metal, then electrons get enough energy to just escape the metal surface._{o}__Case-3__- When (*hv > Φ*),e., energy of photon is greater than the work function of metal. Then electron, apart from getting energy to escape the metal surface, the remaining energy is provided to the electron as kinetic energy. Mathematically, it can be expressed as:_{o}

*hv = Φ _{o} + K_{max}*

Here, *K _{max }*is the maximum kinetic energy of a photoelectron

- The above equation is known as
__Einstein’s photoelectric equation__

From the Einstein’s photoelectric equation, following points are clear:

- Photoelectric current (
*i*) is directly proportional to the intensity (*I*) of radiation. As the intensity rises, number of photons received by metal surface in a unit area per unit time rises, so number of electrons emitted rises, and hence, photoelectric current increases

- Since saturation current is just a maximum value of photoelectric current, saturation current gets higher with increasing intensity of incident light
- For every metal, there exists a certain minimum frequency below which no photoelectric effect occurs. This frequency is called threshold frequency.

Here, *v _{o }*

- Stopping potential (
*V*) and Maximum kinetic energy (_{o}*K*) doesn’t depend upon the intensity. Because intensity is the number of photons in unit area and unit time, and the photoelectric effect take place when one electron takes one photon_{max} - Stopping potential (
*V*) and Maximum kinetic energy (_{o}*K*) is directly proportional to the frequency (_{max}*ν*)

The relation between stopping potential, maximum kinetic energy and the frequency of incident light could be expressed mathematically as follows: using Einstein’s photoelectric equation

**hv = Φ _{o} + K_{max}**

**K _{max} = hν - Φ_{o}**

**Also, K _{max} = eV_{o}**

**∴ eV _{o} = hν - Φ_{o}**

On rearranging the above equation:

**V _{o} = (h/e)v + (Φ_{o}/e)**

Plotting the above equation graphically, we get:

- Photoelectric emission is an
__instantaneous process__, meaning that there is no time gap between incident radiation and electron emission.

__Numerical Problems: __

1)**Question**: Caesium metal has work function of 2.14eV. Photoelectric emission takes place when a light of frequency 6×10^{14 }Hz is incident on the metal surface. Calculate the following: a) maximum kinetic energy of the electrons emitted, b) stopping potential, and c) maximum speed of the emitted electrons.

* Solution*:

Given, Φ_{o} = 2.14eV = 2.14×10^{-19}J, and ν = 6×10^{14}Hz

a) Using Einstein’s photoelectric equation:

*hv = Φ _{o}+ K_{max}*

*K _{max} = hv - Φ_{o} = (6.6×10^{-34}×6×10^{14})J – (2.14×10^{-19})J*

*∴**K _{max} = 1.82×10^{-19}J (ans)*

b) Stopping potential is given by the equation:

*eV _{o} = K_{max}*

*V _{o} = K_{max}/e = 1.82×10^{-19} = 1.1375V (ans)*

c) Maximum speed of emitted electrons can be found using maximum kinetic energy equation:

*K _{max} = (1/2)mv_{max}^{2}*

^{}

2)**Question**: Light of wavelength 488nm is incident on an emitter plate. The photoelectrons have a stopping (cut-off) potential of 0.38V. Calculate the work function of the emitter plate.

* Solution*:

Given, *λ = 488×10 ^{-9}*m,

Using Einstein’s photoelectric equation:

*hv = hc/λ = Φ _{o} + K_{max}*

and,

*K _{max} = eV_{o}*

_{}

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