In an experiment on the photoelectric effect with a given metal, the stopping potential is 0.60 V for light of frequency \( 6.0\times10^{14}\,\mathrm{Hz} \) and 2.25 V for light of frequency \( 1.0\times10^{15}\,\mathrm{Hz} \). The value of Planck’s constant obtained from these readings is (Take \( e=1.6\times10^{-19}\,\mathrm{C} \))
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( 6.6\times10^{-34}\,\mathrm{J\,s} \)
Writing Einstein’s photoelectric equation in terms of the stopping potential gives \( V_0=(h/e)\nu-\phi_0/e \). The graph of \( V_0 \) against \( \nu \) is therefore a straight line of slope \( h/e \). Two readings on this line give the slope, and multiplying the slope by \( e \) gives Planck’s constant.
The slope of the line through the two readings is:
\( \begin{aligned} \frac{h}{e} &= \frac{\Delta V_0}{\Delta\nu} \\ &= \frac{(2.25-0.60)\,\mathrm{V}}{(1.0\times10^{15}-6.0\times10^{14})\,\mathrm{Hz}} \\ &= \frac{1.65\,\mathrm{V}}{4.0\times10^{14}\,\mathrm{Hz}} \\ &= 4.125\times10^{-15}\,\mathrm{V\,s} \end{aligned} \)
Planck’s constant is then:
\( \begin{aligned} h &= e\times\frac{\Delta V_0}{\Delta\nu} \\ &= 1.6\times10^{-19}\,\mathrm{C}\times4.125\times10^{-15}\,\mathrm{V\,s} \\ &= 6.6\times10^{-34}\,\mathrm{J\,s} \end{aligned} \)
NCERT (2026-27):
Einstein’s Photoelectric Equation: Energy Quantum of Radiation – “… or
\( \displaystyle V_0=\left(\frac{h}{e}\right)\nu-\frac{\phi_0}{e} \qquad (11.4) \)
This is an important result. It predicts that the V0 versus ν curve is a straight line with slope = (h/e), independent of the nature of the material. During 1906-1916, Millikan performed a series of experiments on photoelectric effect, aimed at disproving Einstein’s photoelectric equation. He measured the slope of the straight line obtained for sodium …. Using the known value of e, he determined the value of Planck’s constant h.”