Question

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?

Short answer

The required current of electrons ejected from a surface is$9.68\times {10}^{-20}A.$

Step-by-step solution

Identification of the given data:

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 laser output:

Theoutput 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" $\lambda =600nm$

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

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

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

$$\begin{gathered} \mathrm{R}\text{'}=1.0\times {10}^{-16}\mathrm{R} \\ =\left(1.0\times {10}^{-16}\right)\left(6.04\times {10}^{15}\mathrm{photons}/\mathrm{s}\right) \\ =0.604\mathrm{electron}/\mathrm{s}. \end{gathered}$$

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

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

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