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Question
Light of intensity 10−5 W m−2 falls on a sodium photo-cell of surface area 2 cm2. 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?
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Solution
Intensity of incident light, I = 10−5 W m−2
Surface area of a sodium photocell, A = 2 cm2 = 2 × 10−4 m2
Incident power of the light, P = I × A
= 10−5 × 2 × 10−4
= 2 × 10−9 W
Work function of the metal, `phi_0` = 2 eV
= 2 × 1.6 × 10−19
= 3.2 × 10−19 J
Number of layers of sodium that absorbs the incident energy, n = 5
We know that the effective atomic area of a sodium atom, Ae is 10−20 m2.
Hence, the number of conduction electrons in n layers is given as:
`"n'" = "n" xx "A"/"A"_"e"`
= `5 xx (2xx10^(-4))/10^(-20)`
= 1017
The incident power is uniformly absorbed by all the electrons continuously. Hence, the amount of energy absorbed per second per electron is:
`"E" = "P"/"n'"`
= `(2xx10^(-9))/10^17`
= 2 × 10−26 J/s
Time required for photoelectric emission:
`"t" = phi_0/"E"`
= `(3.2 xx 10^(-19))/(2xx10^(-26))`
= 1.6 × 107 s ≈ 0.507 years
The time required for the photoelectric emission is nearly half a year, which is not practical. Hence, the wave picture is in disagreement with the given experiment.
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