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Fundamentals Of Physics
Found in: Page 1182

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Short Answer

Suppose the fractional efficiency of a Cesium surface (with work function 1.80 eV ) is 1.0×10-16 ; that is, on average one electron is ejected for every 1016 photons that reach the surface. What would be the current of electrons ejected from such a surface if it were illuminated with 600 nm light from 2.0 mW laser and all the ejected electrons took part in the charge flow?

The required current of electrons ejected from a surface is 9.68×10-20A.

See the step by step solution

Step by Step Solution

Identification of the given data:

The given data is listed below.

The fractional efficiency is 1.0×10-16

The wavelength of the laser is λ=600 nm.

Power, P= 2.0 mW.

The laser output:

The output of the laser is given by


Where R is the rate of the number of photons emitted per unit time of the laser and E is the energy of a single photon.

To determine the current of electrons ejected from the surface

The power output is given by

P = RE ….. (1)

Now, Energy is given by,


Here, h is the Plank’s constant, c is the speed of light and λ is the wavelength.

Substitute localid="1663150706210" E=hcλ for E into equation (1)

localid="1663150711000" P=hcRλ

Solving the above equation and finding the value of R.

localid="1663150715861" R=Pλhc

Consider the given data below.

The wavelength, localid="1663150719899" λ=600 nm

The power, localid="1663150723665" P=2.0 mW

Plank’s constant, localid="1663150727678" h=6.626×10-34J·s

Speed of light, localid="1663150733098" c=2.998×108m/s

Substitute these numerical values

Now, the fractional efficiency of Cesium surface is 1.0×10-16.

Therefore, the rate of photons that actually cause photoelectric emissions is

R'=1.0×10-16 R =1.0×10-166.04×1015photons/s = 0.604 electron/s.

Now, the current is equal to the rate of electrons multiplied by the charge of an electron


Hence, the current of electrons ejected from the surface is 9.68×10-20A.

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