The minimum energy needed to eject an electron from a metal is known as the work function of the metal. Work function is different for different metals.

A photon having an energy at least equal to the work function can eject electron from the metal, frequency of such a photon whose energy is just equal to the work function is called **Threshold Frequency**.

Work Function ; W = h ν_{o} ; Where , ν_{o} is Threshold Frequency

Electrons are, therefore, emitted only if the frequency of the photon is greater than the threshold frequency.

__Kinetic energy of photoelectron :__

In practice, the emitted photo electrons have a range of kinetic energies.

Suppose that, the energy of an incident photon is E and the work function of the metal is W. The maximum K.E. that the photoelectron can have is given by the expression

T_{max} = E − W

If the frequency of the photon is ν and threshold frequency for the metal is ν_{o}, then

T_{max} = h(ν − ν_{o})

For ν < ν_{o }, electron is not ejected.

We can draw two conclusions:

(i) The kinetic energy of the ejected electron depends linearly on the frequency of the photon.

(ii) The number of Photo electrons ejected per unit time depends on the intensity of the light, frequency of light being kept constant

The experimental arrangement for observing the photoelectric effect is shown in the following figure.

All the ejected electrons are not able to leave the surface and hence are not able to reach the anode. With an increase in potential difference across the tube, the number of electrons reaching the anode increases.

When the potential difference across the tube is increased , the Photo current increases and finally reaches a maximum value (Is) which depends on the intensity of light .

__Stopping Potential__

If the polarity of the battery is reversed and the applied potential is gradually increased, the photo-current starts decreasing.

This is because the electrons are retarded, and most of the elecrons are unable to reach the opposite electrode.

It is observed that when the applied retarding potential is increased, the photocurrent eventually becomes zero. This potential is known as the stopping potential and depends only on the material of the photocathode and the frequency of light.

If V_{s }be the stopping potential then

eV_{s} = hν − W

The stopping potential Vs depends only on the metal and does not depend on the intensity of incident light. a , b , c – different intensites.

Example : Light from a discharge tube containing hydrogen atoms falls on the surface of a piece of sodium. The kinetic energy of the fastest photo-electrons emitted from sodium is 0.73 eV. The work function for sodium is 1.82 eV. Find

(a) the energy of the photons causing photoelectric emission.

(b) the quantum numbers of the two levels involved in the emission of these photons.

(c) the change in the angular momentum of the hydrogen atom in the above transition, and

(d) the recoil speed of the emitting atom assuming it to be at rest before the transition.

(Ionization potential of hydrogen is 13.6 volt and the mass of the hydrogen atom is 1.67 x 10^{-27} Kg, 1 eV = 1.6 x 10^{-19} J).

Solution :

(a) According to Einstein’s photo-electric equation, the maximum kinetic energy E_{K} of the emitted electrons is given by

where hν is the energy of photons causing the photo-electric emission and W is the work-function of the emitting surface.

Given that,

E_{K} = 0.73 eV and W = 1.82 eV

hν = E_{K} + W

= 0.73 eV + 1.82 eV

= 2.55 eV

(b) These photons (whose energy is 2.55 eV) are by hydrogen atoms.

As (I.E.)_{H} = 13.6 eV,

hence E_{1}^{H}= −(I.E.)_{H}

= −13.6 eV

The energy of higher levels is given by

E_{n}^{H}= E_{1}^{H}/ n^{2}

Hence, E_{2}^{H} = −13.6/4

= −3.4 eV

E_{3}^{H} = −13.6/9

= 1.5 eV

E_{4}^{H} = −13.6/16

= −0.85eV

The energy of the emitted photon is 2.55 eV.

Now E_{4}^{H} − E_{2}^{H}

= − 0.85 eV − (−3.4 eV) = 2.55 eV.

Thus the quantum numbers of two levels involved in the emission of photon of energy 2.55 eV are 4 and 2.

(c) The electron transition causing the emission of photon of energy 2.55 eV is from n = 4 level to n = 2 level.

Now, according to Bohr’s 2nd postulate, the angular momentum of electron in the hydrogen atom is (nh/2π). Thus, the change in angular momentum in the above transition is

ΔL = (4h/2π) − (2h/2π) = h/π

(d) The momentum of the photon emitted from the hydrogen atom

p_{ph} = hν/c = 2.55 x (1.6 x 10^{-19})/ 3 x 10^{8}

= 1.36 x 10^{-27} Kg. m/s

According to the law of conservation of momentum, the recoil momentum of a hydrogen atom will be equal and opposite to the momentum of the emitted photon.

Hence the recoil speed of the atom is

$ \displaystyle v = \frac{|momentum|}{mass} $

$ \displaystyle = \frac{|\vec{P_A}|}{m_A}= \frac{|\vec{P_{ph}}|}{m_A} $

$ \displaystyle = \frac{1.36\times 10^{-27}}{1.67\times 10^{-27}} $

= 0.814 m/s

Exercise : A small plate of a metal (work function = 1.17 eV) is placed at a distance of 2 m from a monochromatic light source of wavelength 4.8 x 10^{-7} m and power 1.0 watt. The light falls normally on the plate. Find the number of photons striking the metal plate per square metre per second. Find maximum velocity of photoelectrons.