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Question: Light of intensity \({10^{ - 5}}W{m^{ - 2}}\) falls on a sodium photocell of surface area \(2c{m^2}\...

Light of intensity 10−5Wm−2{10^{ - 5}}W{m^{ - 2}} falls on a sodium photocell of surface area 2cm22c{m^2}. Assuming that the top 55 layers of sodium absorb the incident energy, estimate the time required for photoelectric emission in the wave picture of radiation. The work function of the metal is given to be about 2eV2eV. What is the implication of your answer? Effective atomic area =10−20m2 = {10^{ - 20}}{m^2}.

Explanation

Solution

This question is based on the incident power and the photoelectric emission. To solve the question some of the known values are given. The development of emission of electrons is understood because of the photoelectric emission.

Formula used:
Incident power,
⇒P=I×A\Rightarrow P = I \times A
Where, PP is the power, II is current, and AA is the surface area.

Complete step by step solution: In this question, we have the current, surface area and the number of layers all is given. And we will find the required time. So first we will see the values which are given.
Firstly we will see the value given in the question,
I=10−5Wm−2I = {10^{ - 5}}W{m^{ - 2}}
A=2×10−4m2A = 2 \times {10^{ - 4}}{m^2}
n=5n = 5
So now we have to calculate the value of tt
⇒ϕ=2eV=2×1.6×10−19J\Rightarrow \phi = 2eV = 2 \times 1.6 \times {10^{ - 19}}J
Since the number of conduction electron per atom in sodium is only one
Therefore the effective atomic area will be,
⇒A=10−20m2\Rightarrow {A^{}} = {10^{ - 20}}{m^2}
For five layers, the Total number of conduction layers will be
⇒5×area of one layereffective atomic area\Rightarrow \dfrac{{5 \times area{\text{ of one layer}}}}{{effective{\text{ atomic area}}}}
Putting the values, we get
⇒5×2×10−410−20\Rightarrow \dfrac{{5 \times 2 \times {{10}^{ - 4}}}}{{{{10}^{ - 20}}}}
And on solving the equation,
⇒1017\Rightarrow {10^{17}}
Incident power,
⇒P=I×A\Rightarrow P = I \times A
Putting the required values, we get
⇒10−5×(2×10−4)\Rightarrow {10^{ - 5}} \times \left( {2 \times {{10}^{ - 4}}} \right)
On further solving it,
⇒2×10−9W\Rightarrow 2 \times {10^{ - 9}}W
Wave picture says that for all the continuous electrons incident power is absorbed uniformly. Hence energy absorbed per second per electron.
⇒E=Incident powerNo. of electrons in five layers\Rightarrow E = \dfrac{{Incident{\text{ power}}}}{{No.{\text{ of electrons in five layers}}}}
Now on substituting the values, we get
⇒2×10−91017\Rightarrow \dfrac{{2 \times {{10}^{ - 9}}}}{{{{10}^{17}}}}
After solving the above equation, we get
⇒2×10−26W\Rightarrow 2 \times 10{}^{ - 26}W

Therefore the total time need for the emission of photoelectrons will be
⇒t=energy required per electron for ejection (ϕ0)energy absorbed per second per atom(E)\Rightarrow t = \dfrac{{energy{\text{ required per electron for ejection }}\left( {{\phi _0}} \right)}}{{energy{\text{ absorbed per second per atom}}\left( E \right)}}
Substituting the values required here, we get
⇒2×1.6×10−192×10−26\Rightarrow \dfrac{{2 \times 1.6 \times {{10}^{ - 19}}}}{{2 \times {{10}^{ - 26}}}}
After solving the above equation,
⇒1.67×107s\Rightarrow 1.67 \times {10^7}s

So for this experiment, wave pictures cannot be applied.

Notes When a small quantity of external energy is applied to the valence electrons, they gain enough energy and break the bonding with the parent atom. These electrons, which move freely inside the metal, are referred to as free electrons. However, these free electrons cannot shake off the surface of a metal. It is as a result of the free electrons within the metals don't have enough energy to flee from metal.