Question

Electromagnetic radiation of 8.12×10¹⁸ Hz frequency is applied on a metal surface and caused electron emission. Determine the work function of the metal if the maximum kinetic energy (
`E_k`) of the emitted electron is 4.16×10⁻¹⁷ J.

Answer 1

The work function ϕ of the metal = 53.4196 x 10⁻¹⁶ J

Step-by-step explanation:

When light is incident on a photoelectric material like metal, photoelectrons are emitted from the surface of the metal. This process is called photoelectric effect.

The relationship between the maximum kinetic energy (
`E_(k)`) of the photoelectrons to the frequency of the absorbed photons (f) and the threshold frequency (f₀) of the photoemissive metal surface is:

`E_(k)` = h(f − f₀)

`E_(k)` = hf - hf₀

E is the energy of the absorbed photons: E = hf

ϕ is the work function of the surface: ϕ = hf₀

`E_(k)` = E - ϕ

Frequency f = 8.12×10¹⁸ Hz

Maximum kinetic energy
`E_(k)` = 4.16×10⁻¹⁷ J

Speed of light c = 3 x 10⁸ m/s

Planck's constant h = 6.63 × 10⁻³⁴ Js

E = hf = 6.63 × 10⁻³⁴ x 8.12×10¹⁸

E = 53.8356 x 10⁻¹⁶ J

from
`E_(k)` = E - ϕ ;

ϕ = E -
`E_(k)`

ϕ = 53.8356 x 10⁻¹⁶ - 4.16×10⁻¹⁷

ϕ = 53.4196 x 10⁻¹⁶ J

The work function of the metal ϕ = 53.4196 x 10⁻¹⁶ J