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Q10P

Expert-verifiedFound in: Page 1181

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 ${\mathbf{1}}{\mathbf{.}}{\mathbf{39}}{\mathbf{}}{\mathit{k}}{\mathit{W}}{\mathbf{/}}{{\mathbf{m}}}^{{\mathbf{2}}}$ . (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 3.61 kW.
- The rate of absorption of photons by panel is $1.00\times {10}^{22}\mathrm{photons}/\mathrm{s}.$
- The required time is 60.2 s.

**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.**

**(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.

**(b)**

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

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

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

**(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.

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

Therefore, the required time is 60.2 s.

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