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In a photoelectric experiment combined with a double-slit setup, the separation between slits is d = 0.24 mm, slit-to-screen distance D = 1.2 m, and the point of observation is y = 1.00 mm above the central maximum. The incident light has wavelength equal to the path difference at that point (i.e., the setup selects UV light such that the path difference equals one wavelength). The work function of the emitter is 2.2 eV. Find the stopping potential V (in volts) needed to stop the photocurrent.

1. 2.0 V
2. 3.0 V
3. 4.0 V
4. 5.0 V

Correct answer: 4.0 V

Solution

Path difference at y: Delta = yd/D = (1.00e-3)(0.24e-3)/1.2 = 2.0e-7 m = 200 nm. This is identified as the wavelength of the incident UV light. Photon energy = hc/lambda = 1240 eV*nm / 200 nm = 6.2 eV. By Einstein's equation: eV_stop = hf - phi = 6.2 - 2.2 = 4.0 eV. So stopping potential = 4.0 V.

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