A small piece of cesium metal ( φ = 1.9 eV ) is positioned 20 cm...

The energy of the incident photons is given by:

$E = \dfrac{hc}{\lambda} = \dfrac{1240 \text{ eV nm}}{400 \text{ nm}} = 3.1 \text{ eV}$

According to Einstein's photoelectric equation, the maximum kinetic energy of the emitted photoelectrons is:

K_\max = E - \phi = 3.1 \text{ eV} - 1.9 \text{ eV} = 1.2 \text{ eV}

The minimum kinetic energy of the emitted photoelectrons is:

K_\min = 0 \text{ eV}

For a large metal plate, the electric field just outside the surface with a local charge density $\sigma$ is given by Gauss's law as:

$E_f = \dfrac{\sigma}{\epsilon_0} = \dfrac{1.0 \times 10^{-9}}{8.85 \times 10^{-12}} \approx 113 \text{ V/m}$

The potential difference between the large metal plate and the cesium piece at a distance $d = 20 \text{ cm} = 0.2 \text{ m}$ is:

$\Delta V = E_f \times d = 113 \times 0.2 = 22.6 \text{ V}$

Since the electric field points away from the positively charged plate, it does positive work on the negatively charged electrons moving towards it. The work done on each photoelectron is:

$W = e \Delta V = 22.6 \text{ eV}$

The kinetic energy of the photoelectrons upon reaching the plate will increase by this amount. Therefore, the minimum and maximum kinetic energies at the plate are:

K'_\min = K_\min + W = 0 \text{ eV} + 22.6 \text{ eV} = 22.6 \text{ eV}

K'_\max = K_\max + W = 1.2 \text{ eV} + 22.6 \text{ eV} = 23.8 \text{ eV}

Answer: $22.6 \text{ eV}$ and $23.8 \text{ eV}$