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A metal surface is exposed first to radiation of wavelength λ1 = 350 nm and then to radiation of wavelength λ2 = 540 nm. The maximum speeds of the emitted photoelectrons in the two situations are observed to be in the ratio 2:1. The work function of the metal, in eV, is approximately: (Take photon energy = 1240/λ, with λ in nm, in eV)

1. 1.8
2. 2.5
3. 5.6
4. 1.4

Correct answer: 1.8

Solution

E1 = 1240/350 = 3.54 eV, E2 = 1240/540 = 2.30 eV. KE ratio = (speed ratio)^2 = 4, so (E1 - W) = 4(E2 - W). Solving: 3W = 4*2.30 - 3.54 = 5.64, W = 1.88 ~ 1.8 eV.

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