Q188 / 200 · MHT CET - Medical 2026 (25 April Shift 1) · Physics

A student plots two different graphs using the same photoelectric experiment data.
1. Maximum kinetic energy ($K_{max}$) against frequency ($\nu$)
2. Stopping potential ($V_o$) against frequency ($\nu$)
Let the slope of graph 1 be $S_1$ and the slope of graph 2 be $S_2$.
What is the relationship between the slopes?
(h = Planck's constant, e = charge on electron)

  • A
    $\dfrac{S_1}{S_2} = h$
  • B
    $S_1 \cdot S_2 = h$
  • Correct answer
    C
    $\dfrac{S_1}{S_2} = e$
  • D
    $S_1 + S_2 = e$

Solution

According to Einstein's photoelectric equation:

$K_{max} = h\nu - \Phi$

Comparing this with the equation of a straight line $y = mx + c$, the slope of the $K_{max}$ versus $\nu$ graph is:

$S_1 = h$

The stopping potential $V_o$ is related to maximum kinetic energy by $K_{max} = eV_o$. Substituting this in the photoelectric equation:

$eV_o = h\nu - \Phi$

$V_o = \dfrac{h}{e}\nu - \dfrac{\Phi}{e}$

Comparing this with the equation of a straight line, the slope of the $V_o$ versus $\nu$ graph is:

$S_2 = \dfrac{h}{e}$

Taking the ratio of the two slopes:

$\dfrac{S_1}{S_2} = \dfrac{h}{\dfrac{h}{e}} = e$

Answer: $\dfrac{S_1}{S_2} = e$

Hence, the correct answer is option: 'C'.

Relevant Questions