Answer
$6$ $photons/s$
Work Step by Step
The energy $E_s$ emitted by the source versus time $t$ is given in Fig. 38-26. From Fig. 38-26, we calculate the rate of the energy emission $(R_{E})$ by the source:
$R_{E}=\frac{E_{s}}{t_{s}}$
or, $R_{E}=\frac{7.2\times 10^{-9}}{2}$ $J/s$
or, $R_{E}=3.6\times 10^{-9}$ $W$
Now, energy of a single photon is expressed as, $E=\frac{ch}{\lambda}$,
in which $c$ is the speed of light in vacuum, $\lambda$ is the wavelength of the light and $h$ is the Planck's constant
In our case, $\lambda=600$ $nm$ $=600\times 10^{-9}$ $m$
Therefore, $E=\frac{3\times 10^{8}\times 6.63\times 10^{-34}}{600\times 10^{-9}}$ $J$
or, $E=3.315\times 10^{-19}$ $J$
$\therefore$ The rate of photon emission is given by
$R_{emit}=\frac{R_{E}}{E}$
or, $R_{emit}=\frac{3.6\times 10^{-9}}{3.315\times 10^{-19}}$ $photons/s$
or, $R_{emit}\approx 1.086\times10^{10}$ $photons/s$
The distance between the source and detector is $12$ $m$. The detector has an absorbing area of $2.00 \times 10^{-6}$ $m^2$ and absorbs $50\%$ of the incident light.
Therefore, the rate of photon absorbed by the detector is given by
$R_{abs}=\frac{50}{100}\times\frac{\text{Absorbing area of the detector}}{4\times\pi\times12^2}\times R_{emit}$
or, $R_{abs}=\frac{1}{2}\times\frac{2.00 \times 10^{-6}}{4\times\pi\times144}\times 1.086\times10^{10}$ $photons/s$
or, $R_{abs}=6$ $photons/s$
$\therefore$ The rate of photon absorbed by the detector is $6$ $photons/s$
