A satellite in Earth orbit maintains a panel of solar cells of area $\mathbf{2}\mathbf{.}\mathbf{60}\mathbf{}{\mathbf{m}}^{\mathbf{2}}$perpendicular to the direction of the Sun’s light rays. The intensity of the light at the panel is $\mathbf{1}\mathbf{.}\mathbf{39}\mathbf{}\mathit{k}\mathit{W}\mathbf{/}{\mathbf{m}}^{\mathbf{2}}$. (a) At what rate does solar energy arrive at the panel? (b) At what rate are solar photons absorbed by the panel? Assume that the solar radiation is monochromatic, with a wavelength of 550 nm, and that all the solar radiation striking the panel is absorbed. (c) How long would it take for a “mole of photons” to be absorbed by the panel?

The rate of arrival of solar energy is given by, P = IA……. (1)

Here, I is the intensity, and A is the area.

The energy Eof a photon of wavelength $\lambda$is given by,

$E=\frac{hc}{\lambda }$……. (2)

Here, h is the Planck’s constant, and c is the speed of light.

(a)

Substitute the below values in eq 1.

$$\begin{gathered} \mathrm{I}=1.39\mathrm{kW}/{\mathrm{m}}^2 \\ \mathrm{A}=2.50{\mathrm{m}}^2 \end{gathered}$$

$$\begin{gathered} \mathrm{P}=\left(1.39\mathrm{kW}/{\mathrm{m}}^2\right)\left(2.60{\mathrm{m}}^2\right) \\ =3.61\mathrm{kW} \end{gathered}$$

Therefore, the rate of arrival of solar energy is 3.61 kW.

(b)

Substitute the below values in eq. 2. to calculate the energy of the photon.

$$\begin{gathered} \mathrm{\lambda }=550\mathrm{nm} \\ \mathrm{J}=6.626\times {10}^{-34}\mathrm{J}.\mathrm{s} \\ \mathrm{c}=3\times {10}^8\mathrm{m}/\mathrm{s} \end{gathered}$$

The expression to calculate the rate at which solar photons are absorbed is given by,

$R=\frac{P}{E}$ ...(3)

Substitute the below values in eq 3.

Therefore, the rate of absorption of photon by panel is$1.00\times {10}^{22}\mathrm{photons}/\mathrm{s}.$

(c)

The expression to calculate the time taken by the mole of photons to be absorbed by the panel is given by,

$t=\frac{N_A}{R}$ ...(4)

Substitute the below values in eq 4.

$$\begin{gathered} \mathrm{R}=1.00\times {10}^{22}\mathrm{photons}/\mathrm{s} \\ {\mathrm{N}}_{\mathrm{A}}=6.02\times {10}^{23}\mathrm{photons} \end{gathered}$$

Therefore, the required time is 60.2 s.