The ejection of the photoelectron from the silver metal in the photoelectric effect exeriment can be stopped by applying the voltage of `0.35 V` when the radiation `256.7 nm` is used. Calculate the work function for silver metal.

Options

A. `4.45 eV`
B. `7.86 eV`
C. `5.36 eV`
D. `6.78 eV`

Answer

A

Explanation

According to the phoeoelectric effect equation,
`hv = hv_(0) +KE_("max")`
`:. Hv_(0) (work function) = hv - KE_("max")`
Find the energy of photon `(hv)` through wavelength:
`hv = (hc)/(lambda) =((6.6 xx10^(-34)Js)(3.0xx10^(8)ms^(-1)))/(256.7xx10^(-9)m)`
`= 0.077xx10^(-17)J = 7.7 xx 10^(-19) J`
`= (7.7xx10^(-19)J) ((1eV)/(1.6xx10^(-19)J))`
`= 4.8 eV`
Find the maximum kinetic enegry through stopping potential:
`KE_(max) = eV_(0) = 0.35eV`
`:.` Work function `= (4.8eV) - (0.35eV)`
`= 4.45 eV`