Question

The ejection of the photoelectron from the silver metal in the photoelectric effect experiment 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.

Answer 1

From the principle of conservation of energy, the energy of an incident photon (E) is equal to the sum of the work function (W0) of radiation and its kinetic energy (K.E) i.e.,
E = W0 + K.E
Energy of incident photon (E) = $\frac{hc}{\lambda}$
Where, c= velocity of radiation
h= Planck's constant
$\lambda$ = wavelength of radiation
Substituting the values in the given expression of E: E = $\frac{(6.626 × 10 ^{- 34} Js)(3.0 ×10^8 ms^{ - 1})}{256.7 × 10 ^{-9} m}$ = 7.744 × 10-19 J = $\frac{7.744×10^{-19}}{1.602\times10^{-19}}$ eV
E = 4.83 eV
The potential applied to silver metal changes to kinetic energy (K.E) of the photoelectron. Hence, K.E = 0.35 V
K.E = 0.35 eV
∴ Work function, W0= E - K.E = 4.83 eV - 0.35 eV = 4.48 eV
$\frac{5\lambda_0}{4\lambda_0}=(\frac{5.35}{2.55})^2=\frac{28.6225}{6.5025}$
$\frac{5\lambda_0-2000}{4\lambda_0-2000}$= 4.40177
17.6070 λ0 - 5 λ0 = 8803.537 - 2000
λ0 = $\frac{6805.537}{12.607}$
λ0 = 539.8 nm
λ0 ≈ 540 nm