Question:
In photoelectric effect the graph of stopping potential ($V_0$) versus frequency ($\nu$) is a straight line. The slope of this graph is ______.
In photoelectric effect the graph of stopping potential ($V_0$) versus frequency ($\nu$) is a straight line. The slope of this graph is ______.
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The slope of this graph is a \textbf{Universal Constant}. It does not matter what metal you use for the cathode; the slope ($h/e$) will always be the same.
Updated On: Mar 29, 2026
- $h$
- $\frac{e}{h}$
- $\frac{V_0}{e}$
- $\frac{h}{e}$
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The Correct Option is D
Solution and Explanation
Step 1: Understanding the Concept:
Einstein's photoelectric equation relates the energy of the incident photon to the work function and the maximum kinetic energy of the electron: $h\nu = \phi_0 + K_{max}$.
Step 2: Detailed Explanation:
Since $K_{max} = eV_0$, we can rewrite the equation as: $$eV_0 = h\nu - \phi_0 \implies V_0 = \left(\frac{h}{e}\right)\nu - \frac{\phi_0}{e}$$ This is in the form of a straight line equation $y = mx + c$, where the slope $m = \frac{h}{e}$.
Step 3: Final Answer:
The slope is $h/e$.
Einstein's photoelectric equation relates the energy of the incident photon to the work function and the maximum kinetic energy of the electron: $h\nu = \phi_0 + K_{max}$.
Step 2: Detailed Explanation:
Since $K_{max} = eV_0$, we can rewrite the equation as: $$eV_0 = h\nu - \phi_0 \implies V_0 = \left(\frac{h}{e}\right)\nu - \frac{\phi_0}{e}$$ This is in the form of a straight line equation $y = mx + c$, where the slope $m = \frac{h}{e}$.
Step 3: Final Answer:
The slope is $h/e$.
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