The work function of caesium metal is 2.14 eV. When light of frequency 6 ×10^{14 }Hz is incident on the metal surface, photoemission of electrons occurs. What is the

**(a)** maximum kinetic energy of the emitted electrons,

**(b) **Stopping potential, and

**(c) **maximum speed of the emitted photoelectrons?

#### Solution

Work function of caesium metal, `phi_o = 2.14eV `

Frequency of light, `v = 6.0 xx 10^14 Hz`

**(a)**The maximum kinetic energy is given by the photoelectric effect as:

`K = hv - phi_o`

Where,

*h* = Planck’s constant = 6.626 × 10^{−34} Js

`:. K =(6.626 xx 10^34 xx 6 xx 10^14)/(1.6 xx 10^(-19)) - 2.14`

`= 2.485 - 2.140 = 0.345 eV`

Hence, the maximum kinetic energy of the emitted electrons is 0.345 eV.

b)

For stopping potential `V_0`, we can write the equation for kinetic energy as:

`K =eV_o`

:. `V_o = K/e`

`= (0.345 xx 1.6 xx 10^(-19))/(1.6 xx 10^(-19)) = 0.345 V`

Hence, the stopping potential of the material is 0.345 V.

**c)**Maximum speed of the emitted photoelectrons = *v*

Hence, the relation for kinetic energy can be written as:

`K = 1/2 mv^2`

Where,

*m* = Mass of an electron = 9.1 × 10^{−31} kg

`v^2 = (2K)/m`

`= (2 xx 0.345 xx 1.6 xx 10^(-19))/(9.1 xx 10^(-31)) = 0.1104 xx 10^12`

`:. v = 3.323 xx 10^5 "m/s" = 332.3 "km/s"`

Hence, the maximum speed of the emitted photoelectrons is 332.3 km/s.