Question

When a 6000 Å light falls on the cathode of a photo cell, photoemission takes place. If a potential of 0.8 V is required to stop emission of electron, then determine the

1. frequency of the light
2. energy of the incident photon
3. work function of the cathode material
4. threshold frequency and
5. net energy of the electron after it leaves the surface.

Answer 1

Given: λ = 6000 × 10⁻¹⁰ m

eV₀ = 0.8 eV

i. v = `"C"/λ`

= `(3 xx 10^8)/(6 xx 10^-7)`

= 0.5 × 10¹⁵

v = 5 × 10¹⁴ Hz

ii. E = hv

= `(6.626 xx 10^-34 xx 5 xx 10^14)/(1.6 xx 10^-19)`

= `53.13/1.6 xx 10^-20 xx 10^19`

E = 2.07 eV

iii. ev₀ = `1/2` mv² = 0.8 eV

W = hv – `1/2` mv²

= (2.07 – 0.8) eV

W = 1.27 eV

iv. W = hv₀

v₀ = `(1.27 xx 1.6 xx 10^-19)/(6.626 xx 10^-34)`

= `2.032/6.626 xx 10^15`

v₀ = 3.06 × 10¹⁴ Hz

v. Net Energy = E = hv – hv₀

E = (2.07 – 1.27) eV

E = 0.8 eV