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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 - Physics

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प्रश्न

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?

संख्यात्मक
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उत्तर

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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