Question

If the frequency of the incident radiation is increased from 4 × 10¹⁵ Hz to 8 × 10¹⁵ Hz, by how much will the stopping potential for a given photosensitive surface go up?

Answer 1

The above equation can be written as 
hv - hv₀ = eV_s
hv₁ - hv₀ = eV₁             ....(i)
hv₂ - hv₀ = eV₂             ....(ii)
Subtracting (i) and (ii)

h( v₂ - v₁ ) = e( V₂ - V₁ )

∴ Change in stopping potential = `(h( v_2 - v_1 ))/e`

= `( 6.6( 8 xx 10^15 - 4 xx 10^15 ))/( 1.6 xx 10^-19 ) xx 10^-34`

= `( 6.6 xx 4 xx 10^15 )/( 1.6 xx 10^-19 ) xx 10^-34`

= `( 6.6 xx 4 )/1.6`

= 16.5 V