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Q20P

Expert-verifiedFound in: Page 1182

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 ${\mathbf{1}}{\mathbf{.}}{\mathbf{0}}{\mathbf{\times}}{{\mathbf{10}}}^{\mathbf{-}\mathbf{16}}$ ; 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 $9.68\times {10}^{-20}A.$

The given data is listed below.

The fractional efficiency is $1.0\times {10}^{-16}$

The wavelength of the laser is $\lambda =600nm.$

Power, P= 2.0 mW.

**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.

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" $h=6.626\times {10}^{-34}J\xb7s$

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

Substitute these numerical values

Now, the fractional efficiency of Cesium surface is $1.0\times {10}^{-16}.$

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

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

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

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

Hence, the current of electrons ejected from the surface is $9.68\times {10}^{-20}A.$

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