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In a photoelectric experiment, light of wavelength lambda1 = 3100 angstrom is used in the first setup and lambda2 = 6200 angstrom in the second setup, both on the same metal X. The ratio of maximum speeds of ejected electrons in the two setups is sqrt(5): 1. Find the work function of metal X.
- 1.5 eV
- 6 eV
- 3 eV
- 9 eV
Correct answer: 1.5 eV
Solution
Energy of photon: E = hc/lambda. E1 = hc/3100A, E2 = hc/6200A = E1/2. Let phi = work function. KE1 = E1 - phi, KE2 = E2 - phi = E1/2 - phi. Given v1/v2 = sqrt(5), so KE1/KE2 = v1²/v2² = 5. So (E1 - phi)/(E1/2 - phi) = 5. Solve: E1 - phi = 5*(E1/2 - phi) = 5*E1/2 - 5*phi. 4*phi = 5*E1/2 - E1 = 3*E1/2. phi = 3*E1/8. E1 = hc/3100A = (6.626e-34*3e8)/(3100e-10) = (1.988e-25)/(3.1e-7) = 6.41e-19 J = 4.0 eV. phi = 3*4.0/8 = 1.5 eV.
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