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

Answer:

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

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