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NCERT Class XII Chapter
Dual Nature of Radiation and Matter
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Question : 30 of 37
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Light of intensity 10−5 W m−210^{-5}\,\text{W}\,\text{m}^{-2} falls on a sodium photo-cell of surface area 2 cm2\text{cm}^2. Assuming that the top 5 layers of sodium absorb the incident energy, estimate time required for photoelectric emission in the wavepicture of radiation. The work function for the metal is given to be about 2 eV. What is the implication of your answer?
Solution:  
Wave picture of radiation state that incident energy is uniformly distributed among all the electrons continuously. Let us first calculate the total number of recipient electrons in 5 layers of sodium.
Consider each sodium atom has one electron free as conduction electron.
Effective atomic area = 10−20 m210^{-20}\,\text{m}^2
Number of conduction electrons in 5 layers
n = 5×area of each layereffective area of each atom\frac{5\times\text{area of each layer}}{\text{effective area of each atom}} or n = 5×2×10−410−20\frac{5\times2\times10^{-4}}{10^{-20}} = 101710^{17}
Incident power = incident intensity × area = 10−5×2×10−410^{-5}\times2\times10^{-4} = 2 × 10−910^{-9} W
As incident energy is equally distributed among all conduction electrons.
Energy to each conduction electron per second = 2×10−91017\frac{2\times10^{-9}}{10^{17}} = 2 × 10−2610^{-26} W
Time required for emission by each electron
t = total work function energyenergy received per second\frac{\text{total work function energy}}{\text{energy received per second}}
t = 2 eV2×10−26 W\frac{2\,\text{eV}}{2\times10^{-26}\,\text{W}} = 2×1.6×10−192×10−26\frac{2\times1.6\times10^{-19}}{2\times10^{-26}} = 1.6 × 10710^{7} s = 0.5 year
where experimental observation shows that emission of photoelectrons is instantaneous ≈ 10−910^{-9} sec
Thus, wave picture fails to explain photoelectric effect.
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