Question

In the arrangement shown in the figure, y = 1.0 mm, d = 0.24 mm and D = 1.2 m. The work function of the material of the emitter is 2.2 eV. Find the stopping potential V needed to stop the photocurrent.

Answer 1

Given:
Fringe width, y = 1 mm$\times$2 = 2 mm
Work function, W₀ = 2.2 eV
D = 1.2 m
d = 0.24 mm
Fringe width,
$y=\frac{\lambda \mathrm{D}}{d}$ ,
where $\lambda$ = wavelength of light
$$\begin{gathered} \therefore \lambda =\frac{2\times {10}^{-3}\times 0.24\times {10}^{-3}}{1.2} \\ =4\times {10}^{-7}\mathrm{m} \end{gathered}$$
$$\begin{gathered} \mathrm{Energy},E=\frac{hc}{\lambda } \\ =\frac{4.14\times {10}^{-15}\times 3\times {10}^8}{4\times {10}^{-7}} \\ =3.105\mathrm{eV} \end{gathered}$$
From Einstein's photoelectric equation,
$e{V}_0=E-W_0$,
where V₀ is the stopping potential and e is charge of electron.
$$\begin{gathered} \therefore e{V}_0=3.105-2.2=0.905\mathrm{eV} \\ {V}_0=\frac{0.905}{1.6\times {10}^{-19}}\times 1.6\times {10}^{-19}\mathrm{V} \\ =0.905\mathrm{V} \end{gathered}$$