Light of wavelength 2200 Å (angstorm) falls on photosensitive plate with work function 4.1 eV. Find energy of photon in eV

Explanation

Wavelength = `lambda = 5000 Å `
`lambda = 5000 xx 10^(-10)m = 5 xx 10^(-7) m`
Work function = `phi_0 = 1.9 eV `
(i) Energy of photon = E= ?
`E = (hc)/(lambda)`
`E = (6.6 xx 10^(-34) xx 3 xx 10^8)/(5 xx 10^(-7))`
`E = (19.8)/5 xx 10^(-19) J`
`E = 3.96 xx 10^(-19)J`
`E = (3.96 xx 10^(-19))/(1.6 xx 10^(-19)) eV`
`E = 2.475 eV`
(ii) Maximum kinetic energy = `K.E.("maxi") = E - phi_0`
`K.E._("maxi") = 2.475 - 1:9`
`K.E._("maxi") = 0.575 eV`
(iii) Stopping potential = `V_s = ?`
We know `e V_s = K.E._("maxi") `
`V_s = (K.E._("maxi"))/(e) = (0.575)/e eV`
`V_s = 0.575 V`