If the work function for the photoelectron emission of a metal is 3 . 75 eV , then the threshold wave length…

Work function of metal $\mathrm{\varphi }$=$3.75 \mathrm{eV}=3.75\times 1.6\times {10}^{-19}\mathrm{V}=6\times {10}^{-19}\mathrm{V}$

Planck's constant $\mathrm{h}=6.63\times {10}^{-34}\mathrm{J}.\mathrm{s}$

Velocity of light $\mathrm{c}=3\times {10}^8 \mathrm{m}/\mathrm{s}$

Wavelength = $\frac{\mathrm{hc}}{\mathrm{\varphi }}=\frac{6.63\times {10}^{-34}\times 3\times {10}^8}{6\times {10}^{-19}}=331.5 \times {10}^{-9}\mathrm{m}=331.5 \mathrm{nm}$

Hence, the threshold wave length of the radiation needed for the ejection of the electron is approximately $330 \mathrm{nm}$.