As an electron escapes, energy equivalent to the work function of the emitter substance is 'given up'.

So the

**photon energy **must be greater than or equal to Φ. If the ‘h f' is greater than Φ then the
electron acquires some kinetic energy after leaving the surface. The maximum kinetic energy of the
ejected photoelectron is given by;

**K.E max = ½ m v2**
**max = h f – Φ ………………(i), where m v**^{2}
max = maximum velocity and mass.

This is the Einstein's photoelectric equation.

If the photon energy is just equivalent to work function then, m v2

max = 0, at this juncture the electron

will not be able to move hence no photoelectric current, giving rise to a condition known as cut-off

**frequency, h fco = Φ………………. (ii)**
Also the p.d required to stop the fastest photoelectron is the cut-off potential, V cowhich is given by E = e V co electron volts, but this energy is the maximum kinetic energy of the photoelectrons and therefore,

** ½ m v2**

max = e V co ………….. (iii).

Combining equations (i), (ii) and (iii), we can write Einstein's photoelectric equation as,
e V co = h f – h fco ………………….. (iv)
**NOTE: -- Equations (i) and (iv) are quite useful in solving problems involving photoelectric effect.**
Examples

1. The cut-off wavelength for a certain material is 3.310 × 10-7 m. What is the cut-off frequency for
the material?

**Solution**
Speed of light ‘c' = 3.0 × 108 m/s. Since f = c / λ, then

f = 3.0 × 108/ 3.310 × 10

^{-7}
= 9.06 × 1014 Hz.

2. The work function of tungsten is 4.52 e V. Find the cut-off potential for photoelectrons when a
tungsten surface is illuminated with radiation of wavelength 2.50 × 10-7 m. (Planck's constant, h
= 6.62 × 10^{-34} Js).

**Solution**
Frequency ‘f' = c / λ = 3.0 × 108/ 2.50 × 10

^{-7}
Energy of photon = h f = 6.62 × 10

^{-34} × (3.0 × 108/ 2.50 × 10

^{-7}) × (1 / 1.6 × 10

^{-19})
= 4.97 eV.

Hence h fco = 4.52 e V. e V co = 4.97 e V - 4.52 e V = 0.45 e V = 7.2 × 10

^{-20} J

V co = 7.2 × 10-20 / 1.6 × 10

^{-19} = 0.45 e V.

3. The threshold frequency for lithium is 5.5 × 1014 Hz. Calculate the work function for lithium. (Take
‘h' = 6.626 × 10^{-34} Js)

**Solution**
Threshold frequency, f o = 5.5 × 1014 Hz, ‘h' = 6.626 × 10

^{-34} Js

Φ = h f = 5.5 × 1014× 6.626 × 10

^{-34} = 36.4 × 10

^{-20}
4. Sodium has a work function of 2.0 e V. Calculate

a) The maximum energy and velocity of the emitted electrons when sodium is
Illuminated by a radiation of wavelength 150 nm.

b) Determine the least frequency of radiation by which electrons are emitted.

(Take ‘h' = 6.626 × 10^{-34} Js, e = 1.6 × 10-19, c = 3.0 × 108 m/s and mass of electron
= 9.1 × 10^{-31} kg).

**Solution**
a) The energy of incident photon is given by h f = c / λ

= (6.626 × 10

^{-34} × 3.0 × 108) / 1.50 × 10

^{-9}
= 1.325 × 10

^{-18 }J