Question:
Which one of the four graphs showing lines P, Q, R and S between maximum kinetic energy (E) and intensity of incident radiation (I) is correct?
Which one of the four graphs showing lines P, Q, R and S between maximum kinetic energy (E) and intensity of incident radiation (I) is correct?
Show Hint
Maximum kinetic energy depends on frequency, not on intensity, so the graph against intensity is horizontal.
Updated On: Oct 1, 2026
- P
- Q
- R
- S
Show Solution
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The Correct Option is A
Solution and Explanation
Step 1: Recall Einstein's equation:
The maximum kinetic energy of photoelectrons is \(E_{max} = h\nu - \phi\). It depends on the frequency \(\nu\) of the light and on the work function \(\phi\) of the metal.
Step 2: Role of intensity:
Intensity only tells us how many photons fall on the metal per second. More photons release more electrons, so the photocurrent grows, but each electron still gets the same energy \(h\nu - \phi\).
Step 3: Read the graphs:
In the figure, Q, R and S are inclined straight lines, which would mean the kinetic energy grows with intensity. P is a horizontal line parallel to the intensity axis.
Step 4: Pick the line:
At fixed frequency, \(E_{max}\) stays constant as I changes, so the graph is horizontal. Lines Q, R and S are wrong because they show \(E_{max}\) rising with I.
Final Answer:
The correct graph is the horizontal line P, option (A). \[ \boxed{\text{P}} \]
The maximum kinetic energy of photoelectrons is \(E_{max} = h\nu - \phi\). It depends on the frequency \(\nu\) of the light and on the work function \(\phi\) of the metal.
Step 2: Role of intensity:
Intensity only tells us how many photons fall on the metal per second. More photons release more electrons, so the photocurrent grows, but each electron still gets the same energy \(h\nu - \phi\).
Step 3: Read the graphs:
In the figure, Q, R and S are inclined straight lines, which would mean the kinetic energy grows with intensity. P is a horizontal line parallel to the intensity axis.
Step 4: Pick the line:
At fixed frequency, \(E_{max}\) stays constant as I changes, so the graph is horizontal. Lines Q, R and S are wrong because they show \(E_{max}\) rising with I.
Final Answer:
The correct graph is the horizontal line P, option (A). \[ \boxed{\text{P}} \]
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