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A photoelectron is emitted normally (perpendicularly) from the positive plate of a parallel plate capacitor. The potential difference across the capacitor is 1 V. The wavelength of incident light is 310 nm and the work function of the plate material is 2.2 eV. Given: mass of electron mₑ = 9 * 10^(-31) kg, h*c = 1240 eV*nm. The electron takes 3 ns to reach the negative plate. Find the distance between the plates in mm.

1. 1 mm
2. 2 mm
3. 0.5 mm
4. 1.5 mm

Correct answer: 1 mm

Solution

Initial KE = hc/lambda - phi = 4 - 2.2 = 1.8 eV. Initial velocity u = sqrt(2*1.8*1.6e-19 / 9e-31). The electric field between plates accelerates the electron from positive to negative plate (force on electron is toward negative plate, same as its initial velocity direction). Using d = u*t + (1/2)*a*t² with t=3 ns gives d approximately 1 mm.

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