A light detector has an absorbing area of $\mathbf{2}\mathbf{.}\mathbf{00}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{6}}\mathbf{}{\mathbf{m}}^{\mathbf{2}}$ and absorbs 50% of the incident light, which is at a wavelength 600nm. The detector faces an isotropic source, 12.0 m from the source. The energy E emitted by the source versus time t is given in Fig. 38-26 ( ${\mathit{E}}_{\mathbf{s}}\mathbf{=}\mathbf{7}\mathbf{.}\mathbf{2}\mathit{n}\mathit{J}$, ${\mathit{t}}_{\mathbf{s}}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{s}$ ). At what rate are photons absorbed by the detector?

The expression of emission rate is given by,

$R=\frac{P}{E_p}$

Here, P is power.

The expression for the frequency of the photon is given by,

$f=\frac{c}{\lambda }$

Here, c is the speed of light and $\lambda$is the wavelength.

The energy of a photon is given by,

E=hf

Here, h is the Planck’s constant.

Combine the above two equations.

$E=\frac{hc}{\lambda }$ ….. (1)

Consider the given data as below.

The wavelength, $\lambda =600\mathrm{nm}$

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

The speed of light,$\mathrm{c}=3\times {10}^8\mathrm{m}/\mathrm{s}$

The time,$\Delta t=2.0\mathrm{s}$

The energy, $\Delta E=7.2\mathrm{nJ}$

Substitute the below values in eq 1.

The wavelength, $\lambda =600\mathrm{nm}$

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

The speed of light, $c=3\times {10}^{8}m/s$

$$\begin{gathered} E=\frac{\left(6.626\times {10}^{-34}\mathrm{J}·\mathrm{s}\right)\left(3\times {10}^{8}m/s\right)}{\left(600\mathrm{nm}\right)\left({\displaystyle \frac{1m}{{10}^9}}\right)} \\ =\left(3.3125\times {10}^{-19}\mathrm{J}\right)\left(\frac{1\mathrm{eV}}{1.6\times {10}^{-19}\mathrm{J}}\right) \\ =2.07\mathrm{eV} \end{gathered}$$

The expression of average power output of the source is given by,

$P_{\mathrm{emit}}=\frac{\Delta E}{\Delta t}$….. (2)

Substitute 7.2 nJ for $\Delta E$, and 2.0 s for $\Delta t$in equation (2).

$$\begin{gathered} P_{\mathrm{emit}}=\frac{7.2\mathrm{nJ}}{2.0\mathrm{s}} \\ =\frac{\left(7.2\mathrm{nJ}\right)\left({\displaystyle \frac{1.0eV}{1.6\times {10}^{-19}}}\right)}{2.0s} \\ ==2.25\times {10}^{10}eV/s \end{gathered}$$

Define the rate at which photons radiate.

$$\begin{gathered} R_{\mathrm{emit}}=\frac{P_{\mathrm{emit}}}{E} \\ =\frac{2.25\times {10}^{10}eV/s}{2.07\mathrm{eV}} \\ =1.09\times {10}^{10}photons/s \end{gathered}$$

Since the source is isotropic, then the flux away is just the power radiated divided by the surface area of the sphere.

$\begin{array}{c}I=\frac{P}{A}\\ =\frac{P_{\mathrm{emit}}}{4\pi r^2}\end{array}$

If the detector absorbs light at 50% efficiency, then the power is,

$$\begin{gathered} P_{\mathrm{abs}}=\left(\frac{50}{100}\right)I{A}_{\mathrm{detect}} \\ {\mathrm{R}}_{\mathrm{abs}}\mathrm{E}=\left(0.5\right)\left(\frac{{\mathrm{P}}_{\mathrm{emit}}}{4{\mathrm{\pi r}}^2}\right){\mathrm{A}}_{\mathrm{detect}} \\ {\mathrm{R}}_{\mathrm{abs}}=\left(0.5\right)\left(\frac{{\mathrm{A}}_{\mathrm{detect}}}{4{\mathrm{\pi r}}^2}\right)\left(\frac{{\mathrm{P}}_{\mathrm{emit}}}{\mathrm{E}}\right) \end{gathered}$$….. (3)

Substitute $2.0\times {10}^{-6}{\mathrm{m}}^2$for $A_{\mathrm{detect}},$12.0m for r, and $1.09\times {10}^{10}photons/s$for $R_{\mathrm{emit}}$in equation (3).

Hence, the rate of absorption of the photons by detector is 6.0 photons/s.