Question

Light of intensity ${10}^{–5}W{/}^2$ falls mon a sodium photocell of surface area $2{\text{cm}}^2$ . 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 $2eV$. What is the implication of your answer?

Answer 1

Step 1: Find the value of incident power.
Formula used: $P=I\times A$
Intensity of incident light, $I={10}^{-5}W/m^2$
Surface area of a sodium photocell,
$A=2{\text{cm}}^2=2\times {10}^{-4}{m}^2$
Incident power of the light,
$P=I\times A$
$P={10}^{-5}\times 2\times {10}^{-4}$
$P=2\times {10}^{-9}W$
Work function of the metal, ${\varphi }_0=2eV$
${\varphi }_0=2\times 1.6\times {10}^{-19}J$
${\varphi }_0=3.2\times {10}^{-19}J$
Step 2: Find the number of conducting electrons.
Formula used: $N=n\times \frac{A}{A_e}$
Number of layers of sodium that absorbs the incident energy, n = 5
We know that the effective atomic area of a sodium atom, $A_e$ is ${10}^{-20}{m}^2$
Hence, the number of conduction electrons in n layers is given by,
$N=n\times \frac{A}{A_e}$
$N=5\times \frac{2\times {10}^{-4}}{{10}^{-20}}$
$N={10}^{17}\text{electrons}$
Considering the wave nature, the incident power is uniformly absorbed by all the electrons continuously.
Step 3: Find the energy absorbed.
Formula Used: $E=\frac{P}{N}$
Hence, the amount of energy absorbed per second per electron is:
$E=\frac{P}{N}$
$E=\frac{2\times {10}^{-9}}{{10}^{17}}$
$E=2\times {10}^{-26}J/s$
Step 4: Find the time required for emission.
Formula Used: $t=\frac{{\varphi }_0}{E}$
Time required for photoelectric emission:
$t=\frac{{\varphi }_0}{E}$
$t=\frac{3.2\times {10}^{-19}}{2\times {10}^{-26}}2\times 10-26$
$t=1.6\times {10}^{7}s$
$t\approx 0.5\text{years}$
Experimentally, photoelectric emission is observed nearly instantaneously (~〖10〗^(–9) 𝑠): Thus, the wave picture is in gross disagreement with experiment. In the photon-picture, 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.
Final Answer: $t\approx 0.5\text{years}$