Question:
In an experiment on photoelectric effect, the stopping potential \(V_{0}\) is plotted against the frequency \(\nu\) of the incident light. Considering \(e\) as the electronic charge, the work function of the photoelectric surface is given by
In an experiment on photoelectric effect, the stopping potential \(V_{0}\) is plotted against the frequency \(\nu\) of the incident light. Considering \(e\) as the electronic charge, the work function of the photoelectric surface is given by
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Write Einstein's equation as a straight line of stopping potential against frequency and see what the intercept on the potential axis represents.
Updated On: Nov 12, 2026
- \(OA\) in eV
- \(OB\) in volt
- multiplication of \((OB)\) and \((e)\) in joules
- The slope of the line AB
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The Correct Option is C
Solution and Explanation
Step 1: Understanding the Concept
Einstein's photoelectric equation gives \(eV_0 = h\nu - \phi\), where \(\phi\) is the work function.
Step 2: Straight Line Form
\(V_0 = \dfrac{h}{e}\nu - \dfrac{\phi}{e}\)
The slope is \(\dfrac{h}{e}\) and the intercept on the \(V_0\) axis is \(-\dfrac{\phi}{e}\).
Step 3: Read the Graph
The line cuts the \(V_0\) axis at B, so \(OB = \dfrac{\phi}{e}\) in magnitude. Point A is the threshold frequency.
Step 4: Work Function
\(\phi = e \times OB\), which is in joules when \(e\) is in coulomb and OB is in volt.
Final Answer:
The work function equals the product of OB and the electronic charge, in joules. \[ \boxed{\text{(C) }\text{multiplication of }(OB)\text{ and }(e)\text{ in joules}} \]
Einstein's photoelectric equation gives \(eV_0 = h\nu - \phi\), where \(\phi\) is the work function.
Step 2: Straight Line Form
\(V_0 = \dfrac{h}{e}\nu - \dfrac{\phi}{e}\)
The slope is \(\dfrac{h}{e}\) and the intercept on the \(V_0\) axis is \(-\dfrac{\phi}{e}\).
Step 3: Read the Graph
The line cuts the \(V_0\) axis at B, so \(OB = \dfrac{\phi}{e}\) in magnitude. Point A is the threshold frequency.
Step 4: Work Function
\(\phi = e \times OB\), which is in joules when \(e\) is in coulomb and OB is in volt.
Final Answer:
The work function equals the product of OB and the electronic charge, in joules. \[ \boxed{\text{(C) }\text{multiplication of }(OB)\text{ and }(e)\text{ in joules}} \]
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