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

Data: `lambda_(0)= 3800 Å= 3.8 xx 10^(-7)m, lambda= 2600 Å= 2.6 xx 10^(-7)m, h= 6.63 xx 10^(-34) J.s, c=3 xx 10^(8) m//s, e=1.6 xx 10^(-19) C`
According to Einstein's photoelectric equation,the maximum kinetic energy of the emitted electrons,
`KE_("max") = hv-hv_(0)`
`=hc(1/lambda-1/lambda_(0))`
`=(6.63 xx 10^(-34))(3 xx 10^(8))(10^(7)/2.6- 10^(7)/3.8)`
`=19.89 xx 10^(-19) (3.8 - 2.6)/(2.6 xx 3.8) = (19.89 xx 1.2 xx 10^(-19))/(2.6 xx 3.8)` J
`=(19.89 xx 1.2 xx 10^(-19))/(2.6 xx 3.8 xx 1.6 xx 10^(-19))`eV
`=(19.89 xx 1.2)/(2.6 xx 3.8 xx 1.6) = 1.510` eV