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Home » Light of intensity falls on a sodium photo-cell of surface area . Assuming that the top 5 layers of sodium absorb the incident energy, estimate the time required forphotoelectric emission in the wave-picture of radiation. The work function for the metal is given to be about What is the implication of your answer?
Light of intensity 10^{-5} \mathrm{~W} \mathrm{~m}^{-2} falls on a sodium photo-cell of surface area 2 \mathrm{~cm}^{2}. Assuming that the top 5 layers of sodium absorb the incident energy, estimate the time required for
photoelectric emission in the wave-picture of radiation. The work function for the metal is given to be about 2 \mathrm{eV} . What is the implication of your answer?
CBSE | Class 12 | Dual Nature of Radiation and Matter | NCERT | Physics
Light of intensity 10^{-5} \mathrm{~W} \mathrm{~m}^{-2} falls on a sodium photo-cell of surface area 2 \mathrm{~cm}^{2}. Assuming that the top 5 layers of sodium absorb the incident energy, estimate the time required for
photoelectric emission in the wave-picture of radiation. The work function for the metal is given to be about 2 \mathrm{eV} . What is the implication of your answer?

Intensity of the light is given as =10^{-5} \mathrm{~W} \mathrm{~m}^{-2}

Surface area of the sodium photocell is given as A=2 \mathrm{~cm}^{2}

Incident power of the light is given by the relation,
P=\mid \times A

=10^{-5} \times 2 \times 10^{-4}

=2 \times 10^{-9} \mathrm{~W}

Work function of the metal is \Phi_{0}=2 \mathrm{eV}

=2 \times 1.6 \times 10^{-19}

=3.2 \times 10^{-19} \mathrm{~J}

n=5 is the number of layers of sodium that absorbs the incident energy.

Atomic area of the sodium atom is A_{e} i.e., 10^{-20} \mathrm{~m}^{2}

As a result, in n layers, the number of conduction electrons is,

n^{\prime}=n \times\left(A / A_{e}\right)

=5 \times\left[\left(2 \times 10^{-4}\right) / 10^{-20}\right]=10^{17}

The incident power is continually absorbed by all electrons. Per electron per second, the quantity of energy absorbed is

\mathrm{E}=\mathrm{P} / \mathrm{n}^{\prime}

=\left(2 \times 10^{-9}\right) / 10^{17}

=2 \times 10^{-26} \mathrm{~J} / \mathrm{s}

The time for photoelectric emission

t=\Phi_{0} / \mathrm{E}

=\left(3.2 \times 10^{-19}\right) /\left(2 \times 10^{-26}\right)=1.6 \times 10^{7} \mathrm{~s} \approx 0.507 years

The photoelectric emission takes approximately half a year to complete. This isn’t feasible. As a result, the wave function does not correspond with the experiment.

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