During a photoelectric experiment, the collector plate is...

The energy of the incident photon is given by:

$E = \dfrac{hc}{\lambda} = \dfrac{1240 \text{ eV nm}}{200 \text{ nm}} = 6.2 \text{ eV}$

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

K_\max = E - \phi = 6.2 \text{ eV} - 4.5 \text{ eV} = 1.7 \text{ eV}

The minimum kinetic energy of the emitted photoelectrons is:

K_\min = 0 \text{ eV}

The collector plate is maintained at a positive potential of $2.0 \text{ V}$ relative to the emitter. Therefore, the electrons are accelerated and gain an additional kinetic energy of $2.0 \text{ eV}$.

The maximum kinetic energy of the photoelectrons upon reaching the collector is:

K'_\max = K_\max + eV = 1.7 \text{ eV} + 2.0 \text{ eV} = 3.7 \text{ eV}

The minimum kinetic energy of the photoelectrons upon reaching the collector is:

K'_\min = K_\min + eV = 0 \text{ eV} + 2.0 \text{ eV} = 2.0 \text{ eV}