-->

Wave Optics

Question

The work function of caesium metal is 2.14 eV. When light of frequency 6 x 10¹⁴Hz is incident on the metal surface, photoemission of electrons occurs. What is the stopping potential?

Solution

Work function of caesium metal = 2.14 eV
frequency of light wave = 6 x 10¹⁴Hz

The stopping potential can be calculated as,

{eV}_{0} = K . E ._{\max}

\Rightarrow

V_{0} = \frac{K . E ._{\max}}{e} = \frac{0.35 eV}{e} = 0.35 V