In a photoelectric experiment with one metal, the stopping potential is 1.0 V for light of frequency \(7.5\times10^{14}\ \mathrm{Hz}\) and 2.0 V for light of frequency \(1.0\times10^{15}\ \mathrm{Hz}\). The threshold frequency of the metal is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(5.0\times10^{14}\ \mathrm{Hz}\)
For a given metal, the stopping potential changes linearly with frequency. The threshold frequency is the frequency at which this straight line gives \( V_0=0 \). The two readings fix the slope of the line.
\( \begin{aligned} \text{slope} &= \frac{2.0-1.0}{1.0\times10^{15}-7.5\times10^{14}} \\ &= \frac{1.0}{2.5\times10^{14}} \\ &= 4.0\times10^{-15}\ \mathrm{V\,s} \end{aligned} \)
From the first reading, \( V_0 \) has to fall by 1.0 V more to reach zero.
\( \begin{aligned} \nu_0 &= \nu_1-\frac{V_1}{\text{slope}} \\ &= 7.5\times10^{14}-\frac{1.0}{4.0\times10^{-15}} \\ &= 7.5\times10^{14}-2.5\times10^{14} \\ &= 5.0\times10^{14}\ \mathrm{Hz} \end{aligned} \)
From the above, \( \nu_0=5.0\times10^{14}\ \mathrm{Hz} \). Light of lower frequency gives no photoelectrons from this metal.
NCERT (2026-27):
Effect of frequency of incident radiation on stopping potential – “… The graph shows that (i) the stopping potential V0 varies linearly with the frequency of incident radiation for a given photosensitive material. (ii) there exists a certain minimum cut-off frequency ν0 for which the stopping potential is zero.”