Monochromatic radiation carrying photons of energy 4.8\text{ eV} illuminates a cesium surf… — JAMB UTME

Monochromatic radiation carrying photons of energy $4.8\text{ eV}$ illuminates a cesium surface inside a photoelectric cell. If the work function of cesium is $2.1\text{ eV}$ , determine the stopping potential, in volts, needed to reduce the photoelectric current to zero.

Answer: 2.7 V

Answer

The stopping potential required to reduce the photoelectric current to zero is

2.7\text{ V}

Einstein's photoelectric equation states that incident photon energy

E

equals the work function

W_0

plus the maximum kinetic energy

K_{\text{max}}

of the photoelectrons (

E = W_0 + K_{\text{max}}

). Rearranging gives

K_{\text{max}} = 4.8\text{ eV} - 2.1\text{ eV} = 2.7\text{ eV}

. Since

K_{\text{max}} = e V_s

, an electron-volt value of kinetic energy numerically equals the stopping potential in volts, giving a stopping potential of

2.7\text{ V}

Step-by-Step Solution

Calculate the maximum kinetic energy (

K_{\text{max}}

) of the emitted photoelectrons.

K_{\text{max}} = E - W_0 = 4.8\text{ eV} - 2.1\text{ eV} = 2.7\text{ eV}

According to Einstein's photoelectric equation, incident photon energy is divided into overcoming the work function of the metal and providing kinetic energy to the liberated electron.

Determine the stopping potential (

V_s

) from the maximum kinetic energy.

V_s = \frac{K_{\text{max}}}{e} = \frac{2.7\text{ eV}}{e} = 2.7\text{ V}

The stopping potential

V_s

is the opposing potential difference needed to stop the fastest moving photoelectrons, defined by

K_{\text{max}} = e V_s

Key Concept

Photoelectric Effect and Work Function

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