Question:
Light of wavelength \(1000\AA\) incidents on a metal surface of work function \(6\,eV\). The maximum kinetic energy of photoelectrons is
Light of wavelength \(1000\AA\) incidents on a metal surface of work function \(6\,eV\). The maximum kinetic energy of photoelectrons is
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Shortcut formula:
\[
E(eV)=\frac{12400}{\lambda(\AA)}
\]
Very useful in photoelectric effect numerical problems.
Updated On: Jun 17, 2026
- \(12.4\,eV\)
- \(6.4\,eV\)
- \(19.2\,eV\)
- \(0\,eV\)
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The Correct Option is B
Solution and Explanation
Concept:
Einstein photoelectric equation:
\[
K_{\max}=h\nu-\phi
\]
Also:
\[
E=\frac{12400}{\lambda(\AA)}\,eV
\]
Step 1: Calculate photon energy. Given: \[ \lambda=1000\AA \] Thus: \[ E=\frac{12400}{1000}=12.4\,eV \]
Step 2: Apply Einstein equation. Work function: \[ \phi=6eV \] Hence: \[ K_{\max}=12.4-6 \] \[ K_{\max}=6.4\,eV \] Therefore: \[ \boxed{6.4\,eV} \]
Step 1: Calculate photon energy. Given: \[ \lambda=1000\AA \] Thus: \[ E=\frac{12400}{1000}=12.4\,eV \]
Step 2: Apply Einstein equation. Work function: \[ \phi=6eV \] Hence: \[ K_{\max}=12.4-6 \] \[ K_{\max}=6.4\,eV \] Therefore: \[ \boxed{6.4\,eV} \]
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