Question

In a photoelectric experiment, the collector plate is at 2.0 V with respect to the emitter plate made of copper (φ = 4.5 eV). The emitter is illuminated by a source of monochromatic light of wavelength 200 nm. Find the minimum and maximum kinetic energy of the photoelectrons reaching the collector.

Answer 1

Given:
Work function of copper, ϕ = 4.5 eV,
Wavelength of monochromatic light, λ = 200 nm
From Einstein's photoelectric equation, kinetic energy,
$$\begin{gathered} K=E-\varphi =\frac{h\mathrm{c}}{\lambda }-\varphi , \\ \text{w}\mathrm{here},h=\mathrm{Planck}\text{'s}\mathrm{constant} \\ \mathrm{c}=\mathrm{speed}\mathrm{of}\mathrm{light} \\ \therefore K=\frac{1242}{200}-4.5 \\ =6.21-4.5=1.71\mathrm{eV} \end{gathered}$$

Thus, at least 1.7 eV is required to stop the electron. Therefore, minimum kinetic energy will be 2 eV.
It is given that electric potential of 2 V is required to accelerate the electron. Therefore, maximum kinetic energy
$=(2+1.7)\mathrm{eV}=3.7\mathrm{eV}$