Question:
Using Einstein's photoelectric equation, the graphical representation between the kinetic energy (E) of emitted Photoelectrons and the frequency of incident radiation (\(ν\)) is show correctly in figure
Using Einstein's photoelectric equation, the graphical representation between the kinetic energy (E) of emitted Photoelectrons and the frequency of incident radiation (\(ν\)) is show correctly in figure
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No electrons are emitted below the threshold frequency; above it, E rises linearly.
Updated On: Oct 1, 2026
- A
- B
- C
- D
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The Correct Option is C
Solution and Explanation
Step 1: Understanding the Concept:
Einstein's equation: \(E_k = h\nu - W_0 = h(\nu - \nu_0)\). For \(\nu < \nu_0\) no photoelectron is emitted, so there is no kinetic energy.
Step 2: Shape of the graph:
The graph is a straight line with slope \(h\) that meets the frequency axis at \(\nu_0\). To the left of \(\nu_0\) there is no emission.
Step 3: Compare the four graphs:
(A) A line through the origin would mean \(W_0 = 0\), which is not true for a metal.
(B) A line with a positive intercept on the E axis would give emission with \(\nu = 0\).
(C) A line that starts from the frequency axis at \(\nu_0\) and rises: matches.
(D) A falling line would mean kinetic energy decreases as frequency rises, which is the opposite of the equation.
Step 4: Final Answer:
Graph (C) is correct.
Final Answer:
A straight line starting at the threshold frequency. \[ \boxed{\text{(C) }\text{Graph (C)}} \]
Einstein's equation: \(E_k = h\nu - W_0 = h(\nu - \nu_0)\). For \(\nu < \nu_0\) no photoelectron is emitted, so there is no kinetic energy.
Step 2: Shape of the graph:
The graph is a straight line with slope \(h\) that meets the frequency axis at \(\nu_0\). To the left of \(\nu_0\) there is no emission.
Step 3: Compare the four graphs:
(A) A line through the origin would mean \(W_0 = 0\), which is not true for a metal.
(B) A line with a positive intercept on the E axis would give emission with \(\nu = 0\).
(C) A line that starts from the frequency axis at \(\nu_0\) and rises: matches.
(D) A falling line would mean kinetic energy decreases as frequency rises, which is the opposite of the equation.
Step 4: Final Answer:
Graph (C) is correct.
Final Answer:
A straight line starting at the threshold frequency. \[ \boxed{\text{(C) }\text{Graph (C)}} \]
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