Graphs were plotted between stopping potential (\( V_o \)) and frequency of incident radiation (\( v \)) for 2 metals namely Cesium (work fn. 2.14 eV) and Aluminium (work fn. 4.28 eV). The ratio of slope of graph of Cesium to the slope of graph for Aluminium is:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 1
Einstein’s photoelectric equation gives \( eV_o = h\nu - \phi_0 \), which can be written as a straight line in \( \nu \).
\( \displaystyle V_o = \frac{h}{e}\,\nu - \frac{\phi_0}{e} \)
The slope is \( h/e \), made up of only universal constants, so it is the same for every metal. The work function enters only the intercept.
\( \displaystyle \frac{\text{slope (Cs)}}{\text{slope (Al)}} = \frac{h/e}{h/e} = 1 \)
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):
Einstein’s Photoelectric Equation: Energy Quantum of Radiation – “… 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.”