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The electric field at a point associated with a light wave is given by E = 200[sin(6 × 10 15 )t + sin(9 × 10 15 )t] Vm –1 Given : h = 4.14 × 10 –15 eVs If this light falls on a metal surface having a work function of 2.50 eV, the maximum kinetic energy of the photoelectrons will be

1. 1.90 eV
2. 3.27 eV
3. 3.60 eV
4. 3.42 eV

Correct answer: 3.42 eV

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