cal velocity using photoelectric effect
Answers
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