Light of wavelength `5,500 overset @A` falls on a sensitive plate with work function 1.7 eV .Find stopping potential.

Explanation

`lambda = 4000 Å = 4000 xx 10^(-10) m `
`lambda = 4 xx 10^(-7) m`
Work function = `phi_0 = 2:30 eV `
Stopping potential = `V_s = ? `
Energy of incident radiation = `E = (hc)/(lambda)`
`E = (6.62 xx 10^(-34) xx 3 xx 10^8)/(4 xx 10^(-7))`
`= (19.86)/(4) xx 10^(-34 + 8 + 7)`
`= 4.965 xx 10^(-19) J`
`E = (4.965 xx 10^(-19))/(1.6 xx 10^(-19)) eV = 3.1 eV`
Now `eV_s = E - phi_0`
`= 3.1 - 2.3`
`eV_s = 0.8 eV`
`V_s = 0.8 ` volt.