Match the LIST-I with LIST-II (In the context of a photo-electric effect experiment)
| LIST-I | LIST-II |
| A. Slope of maximum kinetic energy of the electron and frequency graph | I. h/e |
| B. Slope of stopping potential and frequency graph | II. h |
| C. y-intercept of maximum kinetic energy of the electron and frequency graph | III. work function |
| D. y-intercept of stopping potential and frequency graph | IV. Work function/e |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → A - II, B - I, C - III, D - IV
By Einstein’s photoelectric equation, the maximum kinetic energy and the stopping potential are both straight-line functions of frequency.
\( \displaystyle K_{\max} = h\nu - \phi_0 \)
\( \displaystyle eV_0 = h\nu - \phi_0 \;\Rightarrow\; V_0 = \frac{h}{e}\,\nu - \frac{\phi_0}{e} \)
(A) Slope of maximum kinetic energy of the electron and frequency graph — this matches (II) h. The coefficient of \( \nu \) in the \( K_{\max} \) equation is h.
(B) Slope of stopping potential and frequency graph — this matches (I) h/e. Dividing the equation by e makes the slope h/e.
(C) y-intercept of maximum kinetic energy of the electron and frequency graph — this matches (III) work function. At \( \nu = 0 \) the line gives \( K_{\max} = -\phi_0 \), whose magnitude is the work function.
(D) y-intercept of stopping potential and frequency graph — this matches (IV) Work function/e. At \( \nu = 0 \) the line gives \( V_0 = -\phi_0/e \), whose magnitude is the work function divided by e.
NCERT (2026-27):
Einstein’s Photoelectric Equation: Energy Quantum of Radiation – “… If this quantum of energy absorbed exceeds the minimum energy needed for the electron to escape from the metal surface (work function \( \phi_0 \)), the electron is emitted with maximum kinetic energy
\( \displaystyle K_{max} = h\nu - \phi_0 \qquad (11.2) \)”
Einstein’s Photoelectric Equation: Energy Quantum of Radiation – “… Using Eq. (11.1), the photoelectric equation, Eq. (11.2), can be written as
\( \displaystyle eV_0 = h\nu - \phi_0; \quad \text{for } \nu \geq \nu_0 \)
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.”