91Ó°ÊÓ

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,

$f=\frac{c}{\mathrm{λ}}$

Here, c is the speed of light and $\mathrm{λ}$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.

$E=\frac{hc}{\mathrm{λ}}$ ….. (1)

Consider the given data as below.

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

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

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

The time,$\mathrm{Δ}t=2.0\mathrm{s}$

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

Substitute the below values in eq 1.

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

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

The speed of light, $c=3×{10}^{8}m/s$

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

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

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

Substitute 7.2 nJ for $\mathrm{Δ}E$, and 2.0 s for $\mathrm{Δ}t$in equation (2).

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

Define the rate at which photons radiate.

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

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

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

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

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

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

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