Use Einstein’s photoelectric equation

For photoelectric effect, $h\nu = \phi_0 + K_{\max}$

and at stopping potential $V_0$, $K_{\max} = eV_0$

So, $eV_0 = h\nu - \phi_0$

Rearranging, $V_0 = \frac{h}{e}\nu - \frac{\phi_0}{e}$

Interpret the $V_0$ vs frequency $\nu$ graph

Comparing with straight line form $y = mx + c$, $V_0 = \left(\frac{h}{e}\right)\nu - \frac{\phi_0}{e}$

Hence:

• The graph of $V_0$ vs $\nu$ is linear.
• Slope $= \dfrac{h}{e}$.
• Intercept on $V_0$-axis $= -\dfrac{\phi_0}{e}$.
• Threshold frequency, $\nu_0 = \frac{\phi_0}{h}$

Check each statement

(A) $V_0$ vs $\nu$ is linear.
This is true.

(B) Slope of $V_0$ vs $\nu$ curve $= \dfrac{\phi_0}{h}$.
This is false because $\text{slope} = \frac{h}{e}$ not $\dfrac{\phi_0}{h}$.

(C) $h$ constant is related to the slope of $V_0$ vs $\nu$ line.
This is true, since $\text{slope} = \frac{h}{e}$

(D) The value of electric charge of electron is not required to determine $h$ using the $V_0$ vs $\nu$ curve.
This is false because from slope, $\text{slope} = \frac{h}{e} \implies h = e \times \text{slope}$ so $e$ is required.

(E) The work function can be estimated without knowing the value of $h$.
This is true, because from the threshold frequency $\nu_0$ obtained directly from the graph, $V_0=0 \Rightarrow \nu = \nu_0$ and if work function is expressed in electron-volt, $\phi_0 = eV = h\nu_0$ Also, from the intercept on the voltage axis, $\text{intercept} = -\frac{\phi_0}{e}$ so numerically the work function in eV can be estimated from the graph without separately determining $h$.

Correct set of statements

True statements are: $\boxed{(A), (C), (E)}$

Therefore the correct option is: $\boxed{\text{Option A}}$

Comparison with stored answer

Stored correct answer: A
Derived answer: A
Hence, they agree.