Class 12 Physics Dual Nature Radiation Matter Einsteins Photoelectric Equation

## Einstein’s 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 hv, where 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 energyhv ) is incident on a metal surface of work function(Φo), 3 cases could be possible
• Case-1-When (hv < Φo), i.e., energy of photon is less than the work function of metal , no photoelectric emission occurs
• Case-2- When (hv = Φo), 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.
• Case-3- When (hv > Φo),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:

hv = Φo + Kmax

Here, Kmax 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

## i ∝ I

• 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.

## hvo = Φo

Here, vo = frequency of incident light, Φo = work function of metal

• Stopping potential (Vo) and Maximum kinetic energy (Kmax) 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
• Stopping potential (Vo) and Maximum kinetic energy (Kmax) is directly proportional to the frequency (ν)

## Kmax∝ v   Vo∝ v

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 = Φo + Kmax

Kmax = hν - Φo

Also, Kmax = eVo

∴ eVo = hν - Φo

On rearranging the above equation:

Vo = (h/e)v + (Φo/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 6×1014 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, Φo = 2.14eV = 2.14×10-19J, and ν = 6×1014Hz

a) Using Einstein’s photoelectric equation:

hv = Φo+ Kmax

Kmax = hv - Φo = (6.6×10-34×6×1014)J – (2.14×10-19)J

Kmax = 1.82×10-19J (ans)

b) Stopping potential is given by the equation:

eVo = Kmax

Vo = Kmax/e = 1.82×10-19 = 1.1375V (ans)

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

Kmax = (1/2)mvmax2 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 = 488×10-9m,Vo = 0.38V

Using Einstein’s photoelectric equation:

hv = hc/λ = Φo + Kmax

and,

Kmax = eVo .