Question

cal velocity using photoelectric effect

Answer 1

We know that according to Albert Einstein Nobel Prize winning Research paper on Photoelectric effect that :

The energy of incident radiation = work function of the target material + maximum kinetic energy of photo electrons emitted.

let us 1st calculate the energy of incident light

we know that E = hc/λ

where h is the planks constant = 6.625 × 10⁻³⁴ J-sec and c = velocity of light in vaccum = 3 × 10⁸m/sec and given λ = 4.36 ×  10⁻⁷ m

substituting the above values in E = hc/λ we get :

E = {(6.625 × 10⁻³⁴ J-sec) × (3 × 10⁸m/sec)}/(4.36 ×  10⁻⁷ m)

⇒ E = 4.55 × 10⁻¹⁹ Joules

given that the work function of the material = 1.24 eV and we know that 1eV is equal to 1.6 × 10⁻¹⁹ Joules

⇒ The work function of the material in joules :

= 1.24 × 1.6 × 10⁻¹⁹ Joules = 1.98 × 10⁻¹⁹ Joules

as above mentioned we know that E(incident radiation) = Work function + Kinetic energy.

substituting the respective energies in the above equation we get :

4.55 × 10⁻¹⁹ Joules = 1.98 × 10⁻¹⁹ Joules + 1/2 × mass of electron × v²

⇒ mass of electron × v² = 5.132 × 10⁻¹⁹ Joules

we know that mass of electron = 9.1 × 10⁻³¹ kg

⇒ 9.1 × 10⁻³¹ kg  × v² = 5.132 × 10⁻¹⁹ Joules

⇒ v² = 0.563 × 10¹²

⇒ v = 7.43 × 10⁵ m/sec

⇒ The velocity of photoelectron = 7.43 × 10⁵ m/sec