Question:
A metal has work function \(2.5\,\text{eV}\). If a radiation of frequency \(3.2\times10^{15}\,\text{Hz}\) is incident on this metal surface, then the maximum kinetic energy of ejected photoelectrons is \((h=6.6\times10^{-34}\,\text{J-s})\)
A metal has work function \(2.5\,\text{eV}\). If a radiation of frequency \(3.2\times10^{15}\,\text{Hz}\) is incident on this metal surface, then the maximum kinetic energy of ejected photoelectrons is \((h=6.6\times10^{-34}\,\text{J-s})\)
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In photoelectric effect,
\[
K_{\max}=h\nu-\phi.
\]
Convert photon energy from joule to eV before subtracting the work function if the work function is given in eV.
Updated On: Jun 18, 2026
- \(9.5\,\text{eV}\)
- \(2.5\,\text{eV}\)
- \(10.7\,\text{eV}\)
- \(12.6\,\text{eV}\)
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The Correct Option is C
Solution and Explanation
Step 1: Use Einstein's photoelectric equation.
\[ K_{\max}=h\nu-\phi \] where \(\phi\) is the work function.
Step 2: Calculate photon energy.
\[ h\nu=(6.6\times10^{-34})(3.2\times10^{15}) \] \[ h\nu=21.12\times10^{-19}\,\text{J} \] \[ h\nu=2.112\times10^{-18}\,\text{J} \] Since \[ 1\,\text{eV}=1.6\times10^{-19}\,\text{J}, \] \[ h\nu=\frac{2.112\times10^{-18}}{1.6\times10^{-19}} \] \[ h\nu=13.2\,\text{eV} \]
Step 3: Find maximum kinetic energy.
\[ K_{\max}=13.2-2.5 \] \[ K_{\max}=10.7\,\text{eV} \]
Step 4: Final conclusion.
Therefore, \[ \boxed{10.7\,\text{eV}} \]
\[ K_{\max}=h\nu-\phi \] where \(\phi\) is the work function.
Step 2: Calculate photon energy.
\[ h\nu=(6.6\times10^{-34})(3.2\times10^{15}) \] \[ h\nu=21.12\times10^{-19}\,\text{J} \] \[ h\nu=2.112\times10^{-18}\,\text{J} \] Since \[ 1\,\text{eV}=1.6\times10^{-19}\,\text{J}, \] \[ h\nu=\frac{2.112\times10^{-18}}{1.6\times10^{-19}} \] \[ h\nu=13.2\,\text{eV} \]
Step 3: Find maximum kinetic energy.
\[ K_{\max}=13.2-2.5 \] \[ K_{\max}=10.7\,\text{eV} \]
Step 4: Final conclusion.
Therefore, \[ \boxed{10.7\,\text{eV}} \]
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