Question

Light of intensity ${10}^{-5}W{m}^2$ falls on a sodium photo-cell of surface area 2 cm${}^2$. Assuming that the top 5 layers of sodium absorb the incident energy, estimate time required for photoelectric emission in the wave-picture of radiation. The work function for the metal is given to be about 2 eV. What is the implication of your answer?

Answer 1

Assume one conduction electron per atom. Effective atomic area $\sim {10}^{-20}{m}^2$

Number of the electrons in 5 layers is given as:

$\Rightarrow \frac{5\times \text{area of one layer}}{\text{Total atomic area}}$

$=\frac{5\times 2\times {10}^{-4}{m}^2}{{10}^{-20}{m}^2}={10}^{17}$

Incident power is given by: $P=I\times A$

${10}^{-5}\times 2\times {10}^{-4}{m}^2=2\times {10}^{-9}$

In the wave picture, incident power is uniformly absorbed by all the electrons continuously. Consequently, energy absorbed per second per electron:

$E=\frac{\text{Incident Power}}{\text{No. of electrons}}$

$=\frac{2\times {10}^{-9}}{{10}^{17}}=2\times {10}^{-26}W$

Time required for photoelectric emission: $t=\frac{{\varphi }_o}{E}$
$=\frac{2\times 1.6\times {10}^{-19}}{2\times {10}^{-26}}=1.6\times {10}^{7}s$
which is about 0.5 year.

Implication: Experimentally, photoelectric emission is observed nearly instantaneously (109 s): Thus, the wave picture is in gross disagreement with experiment. In the photon-picture, the energy of the radiation is not continuously shared by all the electrons in the top layers. Rather, energy comes in discontinuous quanta. and absorption of energy does not take place gradually. A photon is either not absorbed, or absorbed by an electron nearly instantly