Question:
Graph between stopping potential of most energetic emitted photoelectrons (\(V_s\)) and frequency (\(\nu\)) of incident radiation on metal is given. Value of slope AB/BC in graph is [where \(h\) = Planck’s constant, \(e\) = electronic charge]:
Graph between stopping potential of most energetic emitted photoelectrons (\(V_s\)) and frequency (\(\nu\)) of incident radiation on metal is given. Value of slope AB/BC in graph is [where \(h\) = Planck’s constant, \(e\) = electronic charge]:
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Slope of stopping potential–frequency graph is universal constant \(h/e\).
Updated On: Mar 24, 2026
- \(h\)
- \(e\)
- \(h/e\)
- \(e/h\)
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The Correct Option is C
Solution and Explanation
Step 1: Photoelectric equation: \[ eV_s=h\nu-\phi \]
Step 2: Slope of \(V_s\) vs \(\nu\): \[ \frac{dV_s}{d\nu}=\frac{h}{e} \]
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