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

Explanation

`lambda = 5000 Å = 5500 xx 10^(-10)m`
`lambda = 5.5 xx 10^(-7) m`
Work function = `phi_0 = 1.46 eV `
Stopping potential = `V_s = ?`
Energy of incident radiation = `E = (hc)/(lambda)`
`E = (6.62 xx 10^(-34) xx 3 xx 10^8)/(5.5 xx 10^(-7))`
`E = (19.86)/(5.5) xx 10^(-34+ 8 + 7)`
`= 3.61 xx 10^(-19) J`
`E = (3.61 xx 10^(-19))/(1.6 xx 10^(-19)) eV = 2.26 eV`
Now `eV_s = E - phi_0 = 2.26 - 1.48`
`eV_s = 0.78 eV`
`V_s = 0.78 eV = 0.8` volt.