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

Light of intensity 10−5Wm−2{{10}^{-5}}W{{m}^{-2}} falls on the sodium photocell of surface area 2cm22c{{m}^{2}} and work function 2eV. Assuming that, only top 5 layers of sodium absorbs the incident energy and effective atomic area of the sodium is 10−20m2{{10}^{-20}}{{m}^{2}} , the time required for photoemission in wave picture of light is nearly
a)1012s10\dfrac{1}{2}s
b)12s\dfrac{1}{2}s
c)1012h10\dfrac{1}{2}h
d)12yr\dfrac{1}{2}yr

Explanation

Solution

Sodium has only one valence electron per atom of sodium. Hence we can imply that only one electron is available per atomic area. Therefore we can obtain the total electrons available in the five layers of sodium that absorbs the incident energy. Further after obtaining the rate at which energy is absorbed by an electron, accordingly we can determine the time for which photoemission in wave picture of light is nearly will occur.
Formula used:
P=I×AP=I\times A
E=PNeE=\dfrac{P}{{{N}_{e}}}
t=W∘Et=\dfrac{{{W}_{\circ }}}{E}

Complete step-by-step solution:
Only one electron is available per atom of sodium and only the top five layers of the photocell absorb energy. If the area of the photocell is ‘A’ and the atomic are is ‘a’ then the number of electrons (Ne{{N}_{e}} )which can absorb energy are equal to,
Ne=5Aa ⇒Ne=5(2×10−4)10−20=1017 \begin{aligned} & {{N}_{e}}=\dfrac{5A}{a} \\\ & \Rightarrow {{N}_{e}}=\dfrac{5(2\times {{10}^{-4}})}{{{10}^{-20}}}={{10}^{17}} \\\ \end{aligned}
The power of the incident radiation absorbed by the surface ‘A’ with intensity ‘I’ is given by,
P=I×A ⇒P=10−5×(2×10−4)W ⇒P=2×10−9W \begin{aligned} & P=I\times A \\\ & \Rightarrow P={{10}^{-5}}\times (2\times {{10}^{-4}})W \\\ & \Rightarrow P=2\times {{10}^{-9}}W \\\ \end{aligned}
The electrons absorb energy uniformly per second. Hence the energy absorbed by single electron from the surface per second is given by,
E=PNe ⇒E=2×10−91017=2×10−26J/s \begin{aligned} & E=\dfrac{P}{{{N}_{e}}} \\\ & \Rightarrow E=\dfrac{2\times {{10}^{-9}}}{{{10}^{17}}}=2\times {{10}^{-26}}J/s \\\ \end{aligned}
The time required for photoemission depends on the work function of the surface of the metal. Once a single electron acquires energy equal to the work function (W∘{{W}_{\circ }} )of the material the time taken for energy to be equal to the work function will give us time required for photoemission in wave picture of light. This is given by,
t=W∘E ⇒t=2eV2×10−26J/s=2×1.6×10−192×10−26s ⇒t=1.6×107s≈12yr \begin{aligned} & t=\dfrac{{{W}_{\circ }}}{E} \\\ & \Rightarrow t=\dfrac{2eV}{2\times {{10}^{-26}}J/s}=\dfrac{2\times 1.6\times {{10}^{-19}}}{2\times {{10}^{-26}}}s \\\ & \Rightarrow t=1.6\times {{10}^{7}}s\approx \dfrac{1}{2}yr \\\ \end{aligned}
Therefore the correct answer of the above question is option d.

Note: The work function of a material is defined as the minimum energy required by the electron to escape the metal surface. Different materials have different wave functions. Wave function is an intrinsic property of a surface and does not depend on the incident radiation.