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NTA Abhyas JEE Main2020ChemistryStructure of AtomPractice

A metal surface is illuminated by light of two different wavelengths 248 nm and 310 nm . The maximum speeds of the photoelectrons corresponding to these wavelengths are u 1 and u 2 respectively. If the ratio u 1 u 2 = 2 1 and hc = 1240 eVnm , the work function of the metal is nearly

Options

  1. A3.7 eV
  2. B3.2 eV
  3. C2.8 eV
  4. D2.5 eV

Correct answer

A. 3.7 eV

Step-by-step solution

As K . E ∝ v 2 ratio of kinetic energy = 4 1 , as speed are in the ratio = 2 1 K.E = Energy of incident photon – Work function (W) 4 K E = 1240 248 - W .....(i) and K E = 1240 310 - W .....(ii) By solving (i) and (ii) W = 3.7 e V

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