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Expert-verified Found in: Page 1182 ### Fundamentals Of Physics

Book edition 10th Edition
Author(s) David Halliday
Pages 1328 pages
ISBN 9781118230718 # Suppose the fractional efficiency of a Cesium surface (with work function 1.80 eV ) is ${\mathbf{1}}{\mathbf{.}}{\mathbf{0}}{\mathbf{×}}{{\mathbf{10}}}^{\mathbf{-}\mathbf{16}}$ ; that is, on average one electron is ejected for every ${{10}}^{{16}}$ 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 $9.68×{10}^{-20}A.$

See the 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 $\lambda =600nm.$

Power, P= 2.0 mW.

## The laser output:

The output of the laser is given by

P=RE

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,

$E=\frac{hc}{\lambda }$

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

Substitute localid="1663150706210" $E=\frac{hc}{\lambda }$ for E into equation (1)

localid="1663150711000" $P=\frac{hcR}{\lambda }$

Solving the above equation and finding the value of R.

localid="1663150715861" $R=\frac{P\lambda }{hc}$

Consider the given data below.

The wavelength, localid="1663150719899" $\lambda =600nm\phantom{\rule{0ex}{0ex}}$

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

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

Speed of light, localid="1663150733098" $c=2.998×{10}^{8}m/s$

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

$\mathrm{R}\text{'}=1.0×{10}^{-16}\mathrm{R}\phantom{\rule{0ex}{0ex}}=\left(1.0×{10}^{-16}\right)\left(6.04×{10}^{15}\mathrm{photons}/\mathrm{s}\right)\phantom{\rule{0ex}{0ex}}=0.604\mathrm{electron}/\mathrm{s}.$

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

$i=R\text{'}e\phantom{\rule{0ex}{0ex}}=\left(0.604{s}^{-1}\right)\left(1.602×{10}^{-19}photons/s\right)\phantom{\rule{0ex}{0ex}}=9.68×{10}^{-20}A$

Hence, the current of electrons ejected from the surface is $9.68×{10}^{-20}A.$ ### Want to see more solutions like these? 