If the stopping potential $V_0$ is plotted on the y-axis against the f…

Question

If the stopping potential $V_0$ is plotted on the y-axis against the frequency $
u$ of the incident light on the x-axis for a given metal, what does the slope of the resulting straight line equal? (Treat $e$ as the electronic charge magnitude.)

Accepted Answer

C. $h/e$, Planck's constant over electron charge

1. Einstein's photoelectric equation written for the stopping potential is $eV_0 = h
u - \phi_0$.
2. Rearrange to isolate $V_0$: $V_0 = \dfrac{h}{e}\,
u - \dfrac{\phi_0}{e}$.
3. This is a linear equation in $
u$ of the form $y = mx + c$ with $y = V_0$, $x = 
u$.
4. So the slope $m = \dfrac{h}{e}$ (universal — independent of the metal) and the y-intercept $-\phi_0/e$ (metal-dependent).
5. Option A would have units of energy seconds (wrong dimensions for a voltage/frequency slope). Option B is the inverse of the correct ratio. Option D is the y-intercept magnitude, not the slope.
_Source: NCERT Class 12 Physics Part 2, Ch 11, §11.6 (Einstein's photoelectric equation), p. 7 + Fig. 11.5._

Answer

$h$, Planck's constant

Answer

$e/h$, electron-charge over Planck's constant

Answer

$\phi_0$, the metal's work function

If the stopping potential $V_0$ is plotted on the y-axis against the frequency $\nu$ of the incident light on the x-axis for a given metal, what does the slope of the resulting straight line equal? (Treat $e$ as the electronic charge magnitude.)