91Ó°ÊÓ

The given data is listed below.

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

The wavelength of the laser is $\mathrm{λ}=600nm.$

Power, P= 2.0 mW.

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.

The power output is given by

P = RE ….. (1)

Now, Energy is given by,

$E=\frac{hc}{\mathrm{λ}}$

Here, h is the Plank’s constant, c is the speed of light and $\mathrm{λ}$is the wavelength.

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

localid="1663150711000" $P=\frac{hcR}{\mathrm{λ}}$

Solving the above equation and finding the value of R.

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

Consider the given data below.

The wavelength, localid="1663150719899" $\mathrm{λ}=600nm$

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

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

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

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

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