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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

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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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अध्याय 11: Dual Nature of Radiation and Matter - Exercise [पृष्ठ ४१०]

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एनसीईआरटी Physics Part I and II [English] Class 12
अध्याय 11 Dual Nature of Radiation and Matter
Exercise | Q 11.30 | पृष्ठ ४१०

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(b) Use the same formula you employ in (a) to obtain electron speed for an collector potential of 10 MV. Do you see what is wrong? In what way is the formula to be modified?


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λ1 = 3650 Å, λ2 = 4047 Å, λ3 = 4358 Å, λ4 = 5461 Å, λ5 = 6907 Å,

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