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Light of intensity 10^-5 W m^-2 falls on a sodium photocell of surface area 2 cm^2. Assuming that the top 5 layers of sodium absorb the incident energy, estimate time required for photoelectric emission in the wave-picture of radiation.

Q: 11.30 Light of intensity 10^-^5\hspace{1mm}Wm^-^2  falls on a sodium photo-cell of surface area 2\hspace{1mm}cm^2.  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\hspace{1mm}eV. What is the implication of your answer? (Effective atomic area of a sodium atom = 10-20 m2 )

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Intensity of Incident light(I) =10^-^5\hspace{1mm}Wm^-^2

The surface area of the sodium photocell (A)=2 cm2= 2\times10-4 m2

The rate at which energy falls on the photo cell=IA=2\times10-9 W

The rate at which each of the 5 surfaces absorbs energy= IA/5=4\times10-10W

Effective atomic area of a sodium atom (A')= 10-20 m2

The rate at which each sodium atom absorbs energy is R given by

\\R=\frac{IA}{5}\times \frac{A'}{A}\\ R=\frac{10^{-5}\times 10^{-20}}{5}\\ R=2\times 10^{-26}J/s

The time required for photoelectric emission is

\\t=\frac{\phi _{0}}{R}\\ t=\frac{2\times 1.6\times 10^{-19}}{2\times 10^{-26}}\\ t=1.6\times 10^{7}s\\ t\approx 0.507 \ years

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