A point source is emitting 0.2 W of ultravio- let radiation at a wavelength of `lambda = 2537 Å`. This source is placed at a distance of `1.0 m` from the cathode of a photoelectric cell. The cathode is made of potassium (Work function `= 2.22 eV`) and has a surface area of `4 cm^(2)`.
(a) According to classical theory, what time of exposure to the radiation shall be required for a potassium atom to accumulate sufficient energy to eject a photoelectron. Assume that radius of each potassium atom is 2 Å and it absorbs all energy incident on it.
(b) Photon flux is defined as number of light photons reaching the cathode in unit time. Calculate the photon flux.
(c) Photo efficiency is defined as probability of a photon being successful in knocking out an electron from the metal surface. Calculate the saturation photocurrent in the cell assuming a photo efficiency of 0.1.
(d) Find the cut – off potential difference for the cell.

Options

A. `32.5 nA`
B. `10.15 nA`
C. `65 nA`
D. `3.25 nA`

Answer

A

Explanation

Flux of photon at the cathode `= ((0.1)/(4pi xx 1^(2))) (1/("photon energy")) = 1.015 xx 10^(16) "photons"//"sec" m^(2)`
Saturation current `= ("photon flux" xx "effective area of cathode") xx 5//100 xx 1.6 xx 10^(-19) = 3.25 xx 10^(-8) A`.