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)
Using the cut-off frequencies given in the passage, the difference between the work functions of metals A and B is (Take \(h = 6.6\times10^{-34}\ \mathrm{J\,s}\) and \(1\ \mathrm{eV} = 1.6\times10^{-19}\ \mathrm{J}\).)
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 1.24 eV
The cut-off frequency of a metal is fixed by its work function, \( \nu_0=\phi_0/h \). So each work function is \( \phi_0=h\nu_0 \), and the difference between two work functions is \( h \) times the difference between the cut-off frequencies.
Given:
\( \nu_0=5.0\times10^{14}\ \mathrm{Hz} \)
\( \nu_0^{\prime}=8.0\times10^{14}\ \mathrm{Hz} \)
\( h=6.6\times10^{-34}\ \mathrm{J\,s} \)
\( \begin{aligned} \Delta\phi &= h(\nu_0^{\prime}-\nu_0) \\ &= 6.6\times10^{-34}\times3.0\times10^{14} \\ &= 1.98\times10^{-19}\ \mathrm{J} \end{aligned} \)
In electron volt:
\( \begin{aligned} \Delta\phi &= \frac{1.98\times10^{-19}}{1.6\times10^{-19}}\ \mathrm{eV} \\ &\approx 1.24\ \mathrm{eV} \end{aligned} \)
From the above, the work functions of the two metals differ by about 1.24 eV.
NCERT (2026-27):
Einstein’s Photoelectric Equation: Energy Quantum of Radiation – “… Thus, there exists a threshold frequency ν0 (= φ0/h) for the metal surface, below which no photoelectric emission is possible, no matter how intense the incident radiation may be or how long it falls on the surface.”