Question

The work function of caesium metal is 2.14 eV. When light of frequency 6 × 10¹⁴ 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?

Answer 1

Work function of caesium metal, `phi_"o"` = 2.14 eV

Frequency of light, v = 6.0 × 10¹⁴ 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"_"o"`, 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 = (2"K")/"m"`

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

=  0.1104 × 10¹²

∴ v = 3.323  × 10⁵ m/s

= 332.3 km/s

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