Question

A small piece of cesium metal (φ = 1.9 eV) is kept at a distance of 20 cm from a large metal plate with a charge density of 1.0 × 10⁻⁹ C m⁻² on the surface facing the cesium piece. A monochromatic light of wavelength 400 nm is incident on the cesium piece. Find the minimum and maximum kinetic energy of the photoelectrons reaching the large metal plate. Neglect any change in electric field due to the small piece of cesium present.

Answer 1

Given:
Charge density of the metal plate, $\sigma$ = 1.0 × 10⁻⁹ Cm⁻²
Work function of the cesium metal, φ = 1.9 eV
Wavelength of monochromatic light, $\lambda$ = 400 nm = 400$\times$10^$-9$ m
Distance between the metal plates, d = 20 cm = 0.20 m
Electric potential due to a charged plate,
V = E × d,
where E, the electric field due to the charged plate, is $\frac{\sigma }{{\epsilon }_0}$ and
d is the separation between the plates.
$$\begin{gathered} \therefore V=\frac{\sigma }{{\epsilon }_0}\times d \\ =\frac{1\times {10}^{-9}\times 20}{8.85\times {10}^{-12}\times 100}\left(\because {\epsilon }_0=8.65\times {10}^{-12}{\mathrm{C}}^2{\mathrm{N}}^{-1}-{\mathrm{m}}^{-2}\right) \\ =22.598\mathrm{V}=22.6\mathrm{V} \\ \mathrm{From}\mathrm{Einstein}\text{'s}\mathrm{photoelectric}\mathrm{equation}, \\ e{V}_o=hv-{\mathrm{W}}_0 \\ =\frac{hc}{\lambda }-\mathrm{W} \\ \mathrm{On}\mathrm{substituting}\mathrm{the}\mathrm{respective}\mathrm{values},\mathrm{we}\text{get:} \\ {V}_0=\frac{4.14\times {10}^{-15}\times 3\times {10}^8}{4\times {10}^{-7}}-1.9 \\ =3.105-1.9=1.205\mathrm{eV} \\ =1.205\mathrm{V} \end{gathered}$$

As V₀ is much less than 'V', the minimum energy required to reach the charged plate must be equal to 22.7eV.
For maximum KE, 'V' must have an accelerating value.
Hence maximum kinetic energy,
$$\begin{gathered} \mathrm{K}.\mathrm{E}.={\mathrm{V}}_0+\mathrm{V}=1.205+22.6 \\ =23.8005\mathrm{eV} \end{gathered}$$