A sodium `(Na)` surface is illuminated with light of wavelength `300 nm`. The work function for `Na` metal is `2.46eV`. The kinetic enegry of the ejected photoelectrons is

Options

A. `3.67 eV`
B. `0.68 eV`
C. `1.64 eV`
D. `2.02 eV`

Answer

C

Explanation

Enegry of the incident photon is
`E = hv = (hc)/(lambda) =((6.6 xx10^(-34)Js)(3.0xx10^(8)ms^(-1)))/((300xx10^(-9)m))`
`6.6xx10^(-19)J`
`= (6.6xx10^(-19)J) ((1eV)/(1.6xx10^(-19)J))`
According to the photoelectic effect equation,
`KE_(max) = hv- W_(0)`
`= (4.1eV) - (2.46eV)`
`= 1.64 eV`