A small potassium foil is placed (perpendicular to the direction of incidence of light) a distance r (= 0.5 m) from a point light source whose output power P₀ is 1.0 W. Assuming wave nature of light how long (in s) would it take for the foil to soak up enough energy (= 1.8 eV) from the beam to eject an electron? Assume that the ejected photoelectron collected its energy from a circular area of the foil whose radius equals the radius of a potassium atom (1.3 x 10^-10 m).

Moderate

Solution

If the source radiates uniformly in all directions, the intensity I of the light at a distance r is given by
$I = \frac{P_{0}}{4 π r^{2}} = \frac{1.0 W}{4 π (0.5 m)^{2}} = 0.32 W / m^{2}$
The target area A is n(1.3 x 10^-10 m)² or 5.3 x 10^-20m², so that the rate at which energy falls on the target is given by
$P = I A = (0.32 W / m^{2}) (5.3 \times 10^{- 20} m^{2}) = 1.7 \times 10^{- 20} J / s$
If all this incoming energy is absorbed, the time required to accumulate energy for the electron to escape is
$t = (\frac{1.8 eV}{1.7 \times 10^{- 20} J / s}) (\frac{1.6 \times 10^{- 19}}{1 eV}) = 17 s$

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