Suggested languages for you:

Americas

Europe

Q10P

Expert-verified
Found in: Page 1181

### Fundamentals Of Physics

Book edition 10th Edition
Author(s) David Halliday
Pages 1328 pages
ISBN 9781118230718

# 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 ${\mathbf{1}}{\mathbf{.}}{\mathbf{39}}{\mathbf{}}{\mathbit{k}}{\mathbit{W}}{\mathbf{/}}{{\mathbf{m}}}^{{\mathbf{2}}}$ . (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?

1. The rate of arrival of solar energy is 3.61 kW.
2. The rate of absorption of photons by panel is $1.00×{10}^{22}\mathrm{photons}/\mathrm{s}.$
3. The required time is 60.2 s.
See the step by step solution

## Describe the expression of rate of arrival of solar energy, and energy of the photon

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 E of 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.

## Determine the rate of arrival of solar energy

(a)

Substitute the below values in eq 1.

$\mathrm{I}=1.39\mathrm{kW}/{\mathrm{m}}^{2}\phantom{\rule{0ex}{0ex}}\mathrm{A}=2.50{\mathrm{m}}^{2}$

$\mathrm{P}=\left(1.39\mathrm{kW}/{\mathrm{m}}^{2}\right)\left(2.60{\mathrm{m}}^{2}\right)\phantom{\rule{0ex}{0ex}}=3.61\mathrm{kW}$

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

## Determine the rate of absorption of photon by panel

(b)

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

$\mathrm{\lambda }=550\mathrm{nm}\phantom{\rule{0ex}{0ex}}\mathrm{J}=6.626×{10}^{-34}\mathrm{J}.\mathrm{s}\phantom{\rule{0ex}{0ex}}\mathrm{c}=3×{10}^{8}\mathrm{m}/\mathrm{s}$

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 $1.00×{10}^{22}\mathrm{photons}/\mathrm{s}.$

## Determine the time taken by the mole of photons to be absorbed by the panel

(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.

$\mathrm{R}=1.00×{10}^{22}\mathrm{photons}/\mathrm{s}\phantom{\rule{0ex}{0ex}}{\mathrm{N}}_{\mathrm{A}}=6.02×{10}^{23}\mathrm{photons}$

Therefore, the required time is 60.2 s.