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The work function of caesium is 2.14 eV. Find (a) the threshold frequency for caesium and (b) the wavelength of incident light if the photocurrent is brought to zero by a stopping potential of 0.60 V.

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(a) Let \(v_{0}\) be the threshold frequency. Then \(h v_{0}=w_{0}\)
or \(\quad v_{0}=\frac{\omega_{0}}{\mathrm{~h}}=\frac{2.14 \mathrm{eV}}{6.63 \times 10^{-34}}=5.16 \times 10^{4} \mathrm{~Hz}\)
(b) \(\mathrm{eV}_{\mathrm{s}}=\mathrm{h} v-\mathrm{o}_{0}=\frac{\mathrm{hc}}{\lambda}-\omega_{0}\)
or \(\lambda=\frac{\text { he }}{\mathrm{eV}_{\mathrm{s}}+\mathrm{ea}_{0}}\)
\(\lambda=\frac{6.63 \times 10^{-34} \times 3 \times 10^{8}}{(0.60 \mathrm{eV}+2.14 \mathrm{eV})}=\frac{19.89 \times 10^{-26}}{2.74 \times 1.6 \times 10^{-40}}\) \(=454 \mathrm{~nm}\)
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