The threshold wavelenght of silver is `3800 Å` . Calculate the maximum kinetic energy in eV of photoelelctrons emitted, when ultraviolet light of wavelegth `2600 Å` falls on it. ( Planck's constant, `h=6*63xx10^(-34)"Js",.` Velocity of light in air, `c=3xx10^(8)m//s`)

Explanation

`lamda_(0)=3800 Å`
`lamda_(0)=2660 Å`
`=2*60xx10^(-7)m`
We know that, `K.E_("max")=hc[1/lamda-1/lamda_(m)]`
`K.E_("max")=6*63xx10^(-34)xx3xx10^(8)xx[1/(260xx10^(-7))-1/(3*80xx10^(-7))]`
`=1*989xx10^(-18)[(3*8-2*6)/(3*8xx2*6)]`
`2*4157xx10^(-19)J`
`=(2*4157xx10^(-19))/(1*6xx10^(-19))"eV"`
`= 1*51"eV"`.