Graph shows variation of stopping potential with frequency of incident radiation on a metal plate. The value of Planck's constant is [e = charge on photoelectron]
Show Hint
The slope of a Stopping Potential vs Frequency graph is always $h/e$, regardless of the metal.
Step 1: Understanding the Question:
According to Einstein's photoelectric equation, the stopping potential varies linearly with the frequency of incident light. The slope of this graph is related to Planck's constant $h$. Step 2: Key Formula or Approach:
Einstein's photoelectric equation:
\[ eV_s = hv - \phi_0 \implies V_s = \left(\frac{h}{e}\right)v - \frac{\phi_0}{e} \]
This is in the form $y = mx + c$, where the slope $m = \frac{h}{e}$. Step 3: Detailed Explanation:
From the given graph, consider two points $(v_1, V_1)$ and $(v_2, V_2)$.
The slope of the graph is given by:
\[ \text{Slope} = \frac{\Delta y}{\Delta x} = \frac{V_2 - V_1}{v_2 - v_1} \]
Since $\text{Slope} = \frac{h}{e}$:
\[ \frac{h}{e} = \frac{V_2 - V_1}{v_2 - v_1} \]
\[ h = \frac{e(V_2 - V_1)}{(v_2 - v_1)} \] Step 4: Final Answer:
The value of Planck's constant is $\frac{e(V_2 - V_1)}{(v_2 - v_1)}$.
