A metal surface having a work function `phi = 2.2 xx 10^(-19) J`, is illuminated by the light of wavelengh `1320Å`. What is the maximum kinetic energy ( in eV ) of the emitted photoelectron ? [Take `h = 6.6 xx 10^(-34) Js `]

Answer

8

Explanation

`K. E. = (hc)/(lamda ) - phi _(0)`
`= (6.6 xx 10 ^(-34) xx 3 xx 10 ^(8))/( 1320 xx 10 ^(-10))- 2.2 xx 10 ^(- 19)`
`= 15 xx 10 ^(-19) - 2.2 xx 10 ^(-19)`
`= 12. 8 xx 10 ^(-19) J`
`K. E. = (12. 8 xx 10 ^(-19))/( 1. 6 xx 10 ^(-19)) e V = 8 e V`