Suppose the fractional efficiency of a Cesium surface (with work function 1.80 eV ) is ; that is, on average one electron is ejected for every 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
The given data is listed below.
The fractional efficiency is
The wavelength of the laser is
Power, P= 2.0 mW.
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.
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 is the wavelength.
Substitute localid="1663150706210" for E into equation (1)
Solving the above equation and finding the value of R.
Consider the given data below.
The wavelength, localid="1663150719899"
The power, localid="1663150723665"
Plank’s constant, localid="1663150727678"
Speed of light, localid="1663150733098"
Substitute these numerical values
Now, the fractional efficiency of Cesium surface is
Therefore, the rate of photons that actually cause photoelectric emissions is
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
Question: For the arrangement of Figs. and , electrons in the incident beam in region 1 have energy and the potential step has a height of . What is the angular wave number in (a) region 1 and (b) region 2 ? (c) What is the reflection coefficient? (d) If the incident beam sends electrons against the potential step, approximately how many will be reflected?
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