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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?
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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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When the sun is directly overhead, the surface of the earth receives 1.4 × 103 W m−2 of sunlight. Assume that the light is monochromatic with average wavelength 500 nm and that no light is absorbed in between the sun and the earth's surface. The distance between the sun and the earth is 1.5 × 1011 m. (a) Calculate the number of photons falling per second on each square metre of earth's surface directly below the sun. (b) How many photons are there in each cubic metre near the earth's surface at any instant? (c) How many photons does the sun emit per second?
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The photoelectric current increases with increase of intensity of incident light.
Consider a metal exposed to light of wavelength 600 nm. The maximum energy of the electron doubles when light of wavelength 400 nm is used. Find the work function in eV.
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Why it is the frequency and not the intensity of the light source that determines whether the emission of photoelectrons will occur or not? Explain.
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