A radiation of `5000 overset @ A` is incident on metal surface whose work- function is 1.2 eV. find out the value of stopping potential.

Explanation

`lambda = 5000 Å = 5000 xx 10^(-10) m`
`lambda = 5 xx 10^(-7)m`
Work function `= phi_0 = 1.2 eV`
Stopping potential `= V_s = ?`
`h = 6.6 xx 10^(-34) Js`
`c = 3 xx 10^8 ms^(-1)`
`e = 1.6 xx 10^(-19) C`
Energy of incident radiation= `E= (hc)/(lambda)`
`= (6.6 xx 10^(-34) xx 3 xx 10^8)/(5 xx 10^(-7))`
`= (19.8)/5 xx 10^(-34 + 8 + 7)`
`E = 3.98 xx 10^(-19) J`
`E = (3.98 xx 10^(-19))/(1.6 xx 10^(-19)) eV`
`E = 2.49 eV`
We know `eV_s = E - phi_0`
`eV_s = 2.49 - 1.2`
`eV_s = 1.20 eV`
`V_s = 1.2` volt.