Question:
It takes $4.6$ eV to remove one of the least tightly bound electrons from a metal surface. When monochromatic photons strike the metal surface, electrons having kinetic energy from zero to $2.2$ eV are ejected. What is the energy of the incident photons?
It takes $4.6$ eV to remove one of the least tightly bound electrons from a metal surface. When monochromatic photons strike the metal surface, electrons having kinetic energy from zero to $2.2$ eV are ejected. What is the energy of the incident photons?
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Maximum kinetic energy corresponds to highest emitted electron energy.
Updated On: May 1, 2026
- $2.4$ eV
- $2.2$ eV
- $6.8$ eV
- $4.6$ eV
- $5.8$ eV
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The Correct Option is C
Solution and Explanation
Concept:
Photoelectric equation: \[ E = \phi + K_{max} \]
Step 1: Substitute values.
\[ \phi = 4.6,\quad K_{max} = 2.2 \] \[ E = 4.6 + 2.2 = 6.8 \text{ eV} \]
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