A student plots two different graphs using the same photoelectric...

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