The plot of the stopping potential versus the frequency of the light used in an experiment on the photoelectric effect is a straight line. The stopping potential is $0 \text{ V}$ at a frequency of $1 \times 10^{14} \text{ Hz}$ and $1.656 \text{ V}$ at a frequency of $5 \times 10^{14} \text{ Hz}$ . Determine the work function.

Question

The plot of the stopping potential versus the frequency of the light used in an experiment on the photoelectric effect is a straight line. The stopping potential is

0 \text{ V}

at a frequency of

1 \times 10^{14} \text{ Hz}

and

1.656 \text{ V}

at a frequency of

5 \times 10^{14} \text{ Hz}

. Determine the work function.
The plot of the stopping potential versus the frequency of the light used in an experiment on the photoelectric effect

The plot of the stopping potential versus the frequency of the light used in an experiment on the photoelectric effect

Quantrex Academy · Accepted Answer

Correct answer: C. 0.414 eV

According to Einstein's photoelectric equation:

eV = h\nu - \Phi

From the given graph, the stopping potential is

0 \text{ V}

at the threshold frequency

\nu_0 = 1 \times 10^{14} \text{ Hz}

.

Thus, the work function is

\Phi = h\nu_0 = h(1 \times 10^{14})

.

At

\nu = 5 \times 10^{14} \text{ Hz}

, the stopping potential is

V = 1.656 \text{ V}

. Substituting these values into the equation:

e(1.656) = h(5 \times 10^{14}) - \Phi

Substituting

h(1 \times 10^{14}) = \Phi

into the above equation:

1.656 \text{ eV} = 5\Phi - \Phi

4\Phi = 1.656 \text{ eV}

\Phi = \dfrac{1.656}{4} \text{ eV} = 0.414 \text{ eV}

The plot of the stopping potential versus the frequency of the...

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