Question

The variation of stopping potential for metals A, B, C and D with frequency of incident radiation is as shown in the figure. For which metal, stopping potential is higher for lower values of threshold frequency (\(\nu_0\))? [The frequency of incident radiation \(\nu\) is same.]

• A. C
• B. D
• C. A
• D. B

Hint:

In photoelectric effect, lower work function (lower threshold frequency) means higher kinetic energy of emitted electrons.

Answer 1

The Correct Option is C

Step 1: Photoelectric equation.
Stopping potential \(V_0\) is given by \[ eV_0 = h(\nu - \nu_0). \]
Step 2: Effect of threshold frequency.
For a fixed incident frequency \(\nu\), a lower threshold frequency \(\nu_0\) gives a larger value of \((\nu - \nu_0)\), and hence a larger stopping potential.
Step 3: Interpreting the graph.
From the graph, metal A has the lowest threshold frequency among A, B, C and D.
Step 4: Conclusion.
Therefore, metal A has the highest stopping potential for the given frequency.