Question

The ejection of photoelectrons from a silver metal sample in the photoelectric effect experiment can be stopped by applying a voltage of 0.35 eV with a radiation of 256.7 nm. Calculate the work function for the silver metal sample.

• A. 4.48 eV
• B. 2.48 J
• C. 4.48 J
• D. 2.48 eV

Answer 1

The correct option is A 4.48 eV

Energy of incident radiation
= Work function + K.E of photoelectrons
Energy of incident radiation
$E=h\nu =\frac{hc}{\lambda }=\frac{6.626\times {10}^{-34}Js\times 3.0\times {10}^{8}m{s}^{-1}}{256.7\times {10}^{-9}m}$
$=7.74\times {10}^{-19}J=\frac{7.74\times {10}^{-19}J}{1.602\times {10}^{-19}}=4.83eV$
As stopping potential is given as 0.35 eV
Hence K.E of the electrons = 0.35 eV
Work function = 4.83 eV – 0.35 eV = 4.48 eV.