Question:
The photoemission of electrons occurs when a light of frequency \(5 \times 10^{14}\, Hz\) is incident on a metal surface with work function of \(2.0\, eV\). The maximum speed of emitted photoelectrons is approximately (Planck’s constant \(h = 6.6 \times 10^{-34}\, J\,s\), mass of electron \(m = 9 \times 10^{-31}\, kg\))
The photoemission of electrons occurs when a light of frequency \(5 \times 10^{14}\, Hz\) is incident on a metal surface with work function of \(2.0\, eV\). The maximum speed of emitted photoelectrons is approximately (Planck’s constant \(h = 6.6 \times 10^{-34}\, J\,s\), mass of electron \(m = 9 \times 10^{-31}\, kg\))
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Always convert electron volt into joules before applying \(K = \frac{1}{2}mv^2\) in photoelectric problems.
Updated On: Jul 18, 2026
- \(\frac{\sqrt{5}}{2} \times 10^{5}\, m s^{-1}\)
- \(2\sqrt{3} \times 10^{5}\, m s^{-1}\)
- \(\frac{2\sqrt{5}}{3} \times 10^{5}\, m s^{-1}\)
- \(\frac{2}{\sqrt{3}} \times 10^{5}\, m s^{-1}\)
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The Correct Option is C
Solution and Explanation
Step 1: Use Einstein’s photoelectric equation.
The maximum kinetic energy of emitted photoelectrons is given by: \[ K_{max} = hf - \phi \] where \(hf\) is energy of incident photon and \(\phi\) is work function.
Step 2: Calculate photon energy.
Given: \[ h = 6.6 \times 10^{-34}, \quad f = 5 \times 10^{14} \] \[ hf = 6.6 \times 10^{-34} \times 5 \times 10^{14} = 33 \times 10^{-20} = 3.3 \times 10^{-19}\, J \]
Step 3: Convert work function into joules.
\[ \phi = 2.0\, eV = 2 \times 1.6 \times 10^{-19} = 3.2 \times 10^{-19}\, J \]
Step 4: Compute maximum kinetic energy.
\[ K_{max} = 3.3 \times 10^{-19} - 3.2 \times 10^{-19} = 0.1 \times 10^{-19} = 1 \times 10^{-20}\, J \]
Step 5: Find maximum velocity of electron.
Using: \[ K_{max} = \frac{1}{2}mv^2 \] \[ v = \sqrt{\frac{2K_{max}}{m}} = \sqrt{\frac{2 \times 10^{-20}}{9 \times 10^{-31}}} \] \[ = \sqrt{\frac{2}{9} \times 10^{11}} \]
Step 6: Simplify expression.
\[ v = \sqrt{0.222 \times 10^{11}} = \sqrt{2.22 \times 10^{10}} \approx 1.49 \times 10^{5}\, m s^{-1} \] \[ = \frac{2\sqrt{5}}{3} \times 10^{5}\, m s^{-1} \]
Final conclusion:
\[ \boxed{\frac{2\sqrt{5}}{3} \times 10^{5}\, m s^{-1}} \]
The maximum kinetic energy of emitted photoelectrons is given by: \[ K_{max} = hf - \phi \] where \(hf\) is energy of incident photon and \(\phi\) is work function.
Step 2: Calculate photon energy.
Given: \[ h = 6.6 \times 10^{-34}, \quad f = 5 \times 10^{14} \] \[ hf = 6.6 \times 10^{-34} \times 5 \times 10^{14} = 33 \times 10^{-20} = 3.3 \times 10^{-19}\, J \]
Step 3: Convert work function into joules.
\[ \phi = 2.0\, eV = 2 \times 1.6 \times 10^{-19} = 3.2 \times 10^{-19}\, J \]
Step 4: Compute maximum kinetic energy.
\[ K_{max} = 3.3 \times 10^{-19} - 3.2 \times 10^{-19} = 0.1 \times 10^{-19} = 1 \times 10^{-20}\, J \]
Step 5: Find maximum velocity of electron.
Using: \[ K_{max} = \frac{1}{2}mv^2 \] \[ v = \sqrt{\frac{2K_{max}}{m}} = \sqrt{\frac{2 \times 10^{-20}}{9 \times 10^{-31}}} \] \[ = \sqrt{\frac{2}{9} \times 10^{11}} \]
Step 6: Simplify expression.
\[ v = \sqrt{0.222 \times 10^{11}} = \sqrt{2.22 \times 10^{10}} \approx 1.49 \times 10^{5}\, m s^{-1} \] \[ = \frac{2\sqrt{5}}{3} \times 10^{5}\, m s^{-1} \]
Final conclusion:
\[ \boxed{\frac{2\sqrt{5}}{3} \times 10^{5}\, m s^{-1}} \]
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