The slope of the cut off voltage versus frequency of incident light is found to be \( 4.12\times10^{-15}\,\mathrm{V\,s} \). The value of Planck's constant is:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → e times the slope of cut off voltage versus frequency
By Einstein’s photoelectric equation the maximum kinetic energy of the photoelectrons is \( eV_0 = h\nu - \phi_0 \), where \( V_0 \) is the cut-off (stopping) potential. Writing it as a straight line in \( \nu \):
\( \displaystyle V_0 = \frac{h}{e}\,\nu - \frac{\phi_0}{e} \)
The slope of the \( V_0 \)–\( \nu \) graph is \( \frac{h}{e} \), so \( h = e\times\text{slope} \).
\( h = 1.6\times10^{-19}\,\mathrm{C}\times4.12\times10^{-15}\,\mathrm{V\,s} \approx 6.6\times10^{-34}\,\mathrm{J\,s} \)
Note: This question is from NCERT (2026-27). NTA sets numerical questions from NCERT too, so do not ignore NCERT while practising numericals.
NCERT (2026-27):
Exercises – …
