070421

Question : Figure shows the stopping potential (V₀) for the photoelectron versus 1/λ graph, for two metals A and B, λ being the wavelength of incident light.

a) How is the value of Plank's constant determined from the graph?

b) If the distance between the light source and the surface of metal A is increased, how will the stopping potential from electron emitted from it be affected? Justify your answer.

Doubt by Kulwinder

Solution : 

a) We know, 
According to Einstein's Photoelectric Equation, 
hν=hν₀+Kmax

hν=hν₀+eV₀
hν-hν₀=eV₀
(hν-hν₀)/e=V₀
hν/e-hν₀/e=V₀
hc/λ-hc/λ₀ = V₀
hc/eλ-hc/eλ₀ = V₀

So, 
V₀=hc/eλ-hc/eλ₀
V₀=(hc/e)(1/λ)+(-hc/eλ₀)
On comparing the above equation with y=mx+c

Slope (m) = hc/e

So, 
h=me/c
∴ Plank's Constant (h) = me/c.

b) On increasing the distance between the light source and metal surface, the intensity of the light will change but there is no effect on the frequency of the light source. Since stopping potential depends on frequency of radiation and not on its intensity so we can say that the stopping potential of the emitted electrons will remain same.