Question

A student plots two different graphs using the same photoelectric experiment data.
1. Maximum kinetic energy (\(K_{max}\)) against frequency (\(\nu\))
2. Stopping potential (\(V_{o}\)) against frequency (\(\nu\))
Let the slope of graph 1 be \(S_{1}\) and the slope of graph 2 be \(S_{2}\).
What is the relationship between the slopes?
(h = Planck's constant, e = charge on electron)

• A. \(\frac{S_{1}}{S_{2}}=h\)
• B. \(S_{1}\cdot S_{2}=h\)
• C. \(\frac{S_{1}}{S_{2}}=e\)
• D. \(S_{1}+S_{2}=e\)

Hint:

Write Einstein's equation, then use the relation between maximum kinetic energy and stopping potential to find both slopes.

Answer 1

The Correct Option is C

Step 1: Formula-first approach
Since $V_{o} = \dfrac{K_{max}}{e}$, every value on graph 2 is the value on graph 1 divided by $e$.

Step 2: Calculation
So the slope of graph 2 is the slope of graph 1 divided by $e$: $S_{2} = \dfrac{S_{1}}{e}$.

Step 3: Compare the options
Rearranging gives $\dfrac{S_{1}}{S_{2}} = e$. Also $S_{1}S_{2} = \dfrac{h^{2}}{e}$ and $S_{1}+S_{2}$ has no fixed value, so the other relations fail.

Final Answer:
The slopes are related by S1 divided by S2 equals e. \[ \boxed{\text{(C) }\dfrac{S_{1}}{S_{2}}=e} \]