Question

The work function of a photoelectric material is 4.0 eV. (a) What is the threshold wavelength? (b) Find the wavelength of light for which the stopping potential is 2.5 V.

Answer 1

Work function of a photoelectric material, ϕ = 4 eV = 4 × 1.6 × 10⁻¹⁹ J
Stopping potential, V₀ = 2.5 V
Planck's constant, h = 6.63$\times {10}^{-34}\mathrm{Js}$

(a) Work function of a photoelectric material,
$\varphi =\frac{hc}{{\lambda }_0}$
Here, λ₀ = threshold wavelength of light
c = speed of light
$$\begin{gathered} \therefore {\lambda }_0=\frac{hc}{\varphi } \\ {\lambda }_0=\frac{6.63\times {10}^{-34}\times 3\times {10}^8}{4\times 1.6\times {10}^{-19}} \\ {\lambda }_0=\frac{6.63\times 3}{64}\times \frac{{10}^{-27}}{{10}^{-9}} \\ {\lambda }_0=3.1\times {10}^{-7}\mathrm{m} \\ {\mathrm{\lambda }}_0=310\mathrm{nm} \end{gathered}$$

(b) From Einstein's photoelectric equation,
$$\begin{gathered} E=\varphi +e{V}_0 \\ \mathrm{On}\mathrm{substituting}\mathrm{the}\mathrm{respective}\mathrm{values},\mathrm{we}\text{get:} \\ \frac{hc}{\lambda }=4\times 1.6\times {10}^{-19}+1.6\times {10}^{-19}\times 2.5 \\ \Rightarrow \lambda =\frac{6.63\times {10}^{-34}\times 3\times {10}^8}{6.5\times 1.6\times {10}^{-19}} \\ \Rightarrow \lambda =\frac{6.63\times 3\times {10}^{-26}}{1.6\times {10}^{-19}\times 6.5} \\ \Rightarrow \lambda =1.9125\times {10}^{-7}=191\mathrm{nm} \end{gathered}$$