Read the following passage and answer the questions.
A student uses a photocell to study how the stopping potential \(V_0\) depends on the frequency \(\nu\) of the incident light. She uses one source of light and puts a different filter in front of it for each frequency. The source stays at the same distance from the cell in every run. For each frequency, she records the photocurrent while she changes the potential of the collector plate. Figure (a) shows her results for three frequencies \(\nu_1\), \(\nu_2\) and \(\nu_3\). Each curve meets the potential axis at a different negative potential.
She then repeats the work with emitters of two different metals, A and B. For each metal she plots the stopping potential against the frequency and gets a straight line, as in figure (b). The line for metal A starts at the cut-off frequency \(\nu_0\), and the line for metal B starts at \(\nu_0^{\prime}\). In her experiment, \(\nu_0 = 5.0\times10^{14}\ \mathrm{Hz}\) and \(\nu_0^{\prime} = 8.0\times10^{14}\ \mathrm{Hz}\).
(a)
(b)
Which metal in figure (b) will have the higher stopping potential at a particular frequency \( \nu\gt \nu_0^{\prime}\gt \nu_0 \)?
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → Metal A
The passage says that the stopping potential varies linearly with the frequency. Each metal also has a cut-off frequency at which the stopping potential is zero. In figure (b), the two lines rise with the same slope. The line of metal A starts at the lower cut-off frequency \( \nu_0 \), and the line of metal B at the higher \( \nu_0^{\prime} \). So at any frequency \( \nu \) above \( \nu_0^{\prime} \), the line of metal A lies above the line of metal B.
Option (1) Metal A — Correct. Metal A has the lower cut-off frequency. So less of the photon energy is used to free an electron, and the photoelectrons get a larger maximum kinetic energy. A larger stopping potential is needed to stop them.
Option (2) Metal B — Incorrect. Metal B needs a higher frequency \( \nu_0^{\prime} \) to start emission. So at the same \( \nu \) its electrons come out with less kinetic energy, and its line lies below that of metal A.
Option (3) Both metal A and metal B — Incorrect. The two metals have different cut-off frequencies. So at the same \( \nu \) their stopping potentials are different, and only one of them is higher.