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?

Solution

Given,
Intensity of light = 10^–5 W m^–2

Surface are of the sodium photocell, A = 2 cm²

Top five layers of sodium absorb the incident energy. (given)

work function for the metal,

ϕ_{o}

= 2 eV

Therefore,

Number of atoms in 5 layers of sodium is,

= \frac{5 \times area of each layer}{Effective area of atom} = \frac{5 \times 2 \times 10^{- 4}}{10^{- 20}} = 10^{17}

Assume that there is only one conduction electron per sodium atom.
∴ Number of electrons in 5 layers = 10¹⁷Energy received by an electron per sec is,

= \frac{Power of incident light}{Number of electrons} = \frac{10^{- 5} \times 2 \times 10^{- 4}}{10^{17}} = 2 \times 10^{- 26} W

Time required for photoemission is,

= \frac{Energy required per electron}{Energy absorbed per second per electron} = \frac{2 \times 1.6 \times 10^{- 19}}{2 \times 10^{- 26}} s = 1.6 \times 10^{7} s .

Thus, it is contrary to the observed fact that there is no time lag between the incidence of light and the emission of photoelectrons.

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