Question:medium

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

Step 1: Use the photoelectric equation.
The maximum kinetic energy of an ejected electron is the photon energy minus the work function. \[ K_{max} = E_{photon} - \phi \]

Step 2: Find the photon energy quickly.
For light, photon energy in eV is found from a handy formula. \[ E = \frac{12400}{\lambda(\text{in }\AA)}\,\text{eV} \]
Step 3: Put in the wavelength.
The wavelength is $1000\,\AA$. \[ E = \frac{12400}{1000} = 12.4\,\text{eV} \]
Step 4: Note the work function.
The work function is $\phi = 6$ eV.
Step 5: Subtract to find the kinetic energy.
\[ K_{max} = 12.4 - 6 \]
Step 6: State the answer.
\[ K_{max} = 6.4\,\text{eV} \] \[ \boxed{6.4\,\text{eV}} \]
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