Question

13. The ejection of the photoelectrons 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.

11th - 12th Class

Chemistry

Solution

124

4.0 (1 ratings)

( therefore ) Work function ( = ) Energy of the incident radiation - KE of photoelectron Now energy of the incident radiation ( (mathrm{E})=mathrm{hv} ) [ =frac{h c}{lambda} ] Here ( h=6.626 times 10^{-34} mathrm{Js}: mathrm{c}=3.0 times 10^{8} mathrm{ms}^{-1} ) [ lambda=256.7 mathrm{nm}=256.7 times 10^{-9} mathrm{m} ] Substituting the values in eq. (2), we have ( therefore ) Energy of incident radiation [ begin{aligned} E &=frac{left(6.626 times 10^{-34} mathrm{Js}right) timesleft(3 times 10^{8} mathrm{ms}^{-1}right)}{256.7 times 10^{-9} mathrm{m}} &=7.74 times 10^{-19} mathrm{J} &=4.83 mathrm{eV}left[because 1 mathrm{eV}=1.602 times 10^{-19} mathrm{J}right] end{aligned} ] The potential applied gives the kinetic energy to the electron. Hence, the kinetic energy of the electron ( =4.4 mathrm{eV} ). Substituting the values in eq. (1), we have [ begin{array}{l} text { Work function }=4.83 mathrm{eV}-0.35 mathrm{eV} qquad=4.48 mathrm{eV} end{array} ]