Monochromatic light of frequency 9.0 \times 10^{14}\text{ Hz} is incident on a clean potas… — JAMB UTME

Monochromatic light of frequency $9.0 \times 10^{14}\text{ Hz}$ is incident on a clean potassium emitter surface having a threshold frequency of $5.0 \times 10^{14}\text{ Hz}$ . Given that Planck's constant $h = 6.6 \times 10^{-34}\text{ J}\cdot\text{s}$ and the elementary charge $e = 1.6 \times 10^{-19}\text{ C}$ , what is the stopping potential in volts required to completely arrest the photoelectric current?

Answer: 1.65 V

Answer

The stopping potential required to arrest the emitted photoelectrons is

1.65\text{ V}

Applying Einstein's photoelectric equation

K_{\max} = h f - h f_0 = h(f - f_0)

gives a maximum kinetic energy of

2.64 \times 10^{-19}\text{ J}

. Dividing this by the electronic charge

e = 1.6 \times 10^{-19}\text{ C}

yields the stopping potential

V_s = 1.65\text{ V}

Step-by-Step Solution

Determine the net energy available for photoelectron kinetic energy using Einstein's photoelectric equation

K_{\max} = h(f - f_0) = 6.6 \times 10^{-34}\text{ J}\cdot\text{s} \times (9.0 \times 10^{14} - 5.0 \times 10^{14})\text{ Hz} = 2.64 \times 10^{-19}\text{ J}

Photoelectron emission occurs only when photon energy

hf

exceeds the work function

W_0 = h f_0

, with the excess energy appearing as maximum kinetic energy.

Express stopping potential in terms of electron charge and maximum kinetic energy

V_s = \frac{K_{\max}}{e} = \frac{2.64 \times 10^{-19}\text{ J}}{1.6 \times 10^{-19}\text{ C}} = 1.65\text{ V}

The stopping potential

V_s

does work

e V_s

equal to the maximum kinetic energy

K_{\max}

of the fastest photoelectrons to bring them to rest.

Key Concept

Einstein's Photoelectric Equation and Stopping Potential Relationship

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