Question:

When the wavelength of radiation falling on a metal is changed from 500 nm to 200 nm, the maximum kinetic energy of the photoelectrons becomes three times larger. The work function of the metal is close to :

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Apply \(K_{max} = hc/\lambda - \phi\) for both wavelengths and set \(K_2 = 3K_1\).
Updated On: Nov 5, 2026
  • 1.02 eV
  • 0.81 eV
  • 0.61 eV
  • 0.52 eV
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The Correct Option is C

Solution and Explanation

Use \(hc = 1240\ \mathrm{eV\,nm}\).
Photon energies: \(E_1 = 1240/500 = 2.48\) eV, \(E_2 = 1240/200 = 6.2\) eV.
\(K_1 = 2.48-\phi\), \(K_2 = 6.2-\phi\) and \(K_2 = 3K_1\):
\(6.2-\phi = 3(2.48-\phi) = 7.44-3\phi\)
\(2\phi = 1.24 \Rightarrow \phi = 0.62\) eV, close to 0.61 eV.
Answer: (3) 0.61 eV
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