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

Question

The difference between threshold wavelengths for two metal surfaces $ A $ and $ B $ having work functions $ \phi_A = 9 \, \text{eV} $ and $ \phi_B = 4.5 \, \text{eV} $ is: \[ \text{(Given: } h c = 1242 \, \text{eV nm)} \]

Answer 1

Step 1: Intercept Method
At zero frequency the line gives $V_0 = -\dfrac{\phi}{e}$, which is the point B on the axis.

Step 2: Magnitude
Therefore $\phi = e \times |OB|$.

Step 3: Eliminate Options
OA is a frequency, not an energy. The slope AB equals $\dfrac{h}{e}$, not the work function. Only the product of OB and $e$ gives the work function in joules.

Final Answer:
The work function is $e$ multiplied by OB, in joules. \[ \boxed{\text{(C) }\text{multiplication of }(OB)\text{ and }(e)\text{ in joules}} \]

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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The Correct Option is C

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