Question:
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]
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]
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The slope of a Stopping Potential vs Frequency graph is always $h/e$, regardless of the metal.
Updated On: Apr 30, 2026
- $\frac{e(V_2 - V_1)}{v_1 v_2}$
- $\frac{e V_1 V_2}{(v_2 - v_1)}$
- $\frac{e(V_2 - V_1)}{(v_2 - v_1)}$
- $\frac{e(V_1 v_2)}{(v_2 - v_1)}$
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The Correct Option is C
Solution and Explanation
Step 1: Einstein's Equation
$eV_s = h\nu - \phi \Rightarrow V_s = \frac{h}{e}\nu - \frac{\phi}{e}$
Step 2: Slope Identification
The slope of the $V_s$ vs $\nu$ graph is $m = \frac{h}{e}$.
Step 3: Calculating Slope from Graph
Slope $= \frac{V_2 - V_1}{\nu_2 - \nu_1}$
Step 4: Solving for h
$\frac{h}{e} = \frac{V_2 - V_1}{\nu_2 - \nu_1} \Rightarrow h = \frac{e(V_2 - V_1)}{\nu_2 - \nu_1}$
Final Answer:(C)
$eV_s = h\nu - \phi \Rightarrow V_s = \frac{h}{e}\nu - \frac{\phi}{e}$
Step 2: Slope Identification
The slope of the $V_s$ vs $\nu$ graph is $m = \frac{h}{e}$.
Step 3: Calculating Slope from Graph
Slope $= \frac{V_2 - V_1}{\nu_2 - \nu_1}$
Step 4: Solving for h
$\frac{h}{e} = \frac{V_2 - V_1}{\nu_2 - \nu_1} \Rightarrow h = \frac{e(V_2 - V_1)}{\nu_2 - \nu_1}$
Final Answer:(C)
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