Chapter 27, Problem 86GP

To determine

The number of photons.

Answer to Problem 86GP

  $\text{3}\text{.33}\text{×}{\text{10}}^{\text{52}}\text{Photons/y}$

Explanation of Solution

Given info:

Wavelength, $\text{λ}\text{=}\text{550}\text{nm}\text{=}\text{550}\text{×}{\text{10}}^{\text{-9}}\text{m}$ and

Total power output is $1350{\text{W/m}}^{\text{2}}$ .

The earth-sun distance, radius, $\text{r}\text{=}\text{150}{\text{×10}}^{\text{9}}\text{m}$

Intensity is $\text{2}{\text{.83×10}}^{\text{23}}{\text{m}}^{\text{2}}$

Formula used:

  $\frac{\text{E}}{\text{hf}}\text{=}\frac{\text{Eλ}}{\text{hc}}$

Where,

  $f$ is frequency.

  $E$ is photon’s energy.

  $\lambda$ is wavelength.

  $h$ is Planck’s constant, $6.63\times {10}^{-34}J\cdot s$

  $c$ is speed of light, ${\text{3×10}}^{\text{8}}\text{ m/s }$

Calculation:

First calculate the area of sphere whose radius is the Earth−Sun distance by using the formula, $\text{A}\text{=}{\text{4πr}}^{\text{2}}$

  $\text{A=}\text{4π}{\left({\text{150×10}}^{\text{9}}\text{m}\right)}^{\text{2}}\text{=2}{\text{.83×10}}^{\text{23}}{\text{m}}^{\text{2}}$

Secondly, to calculate the Sun’s total power output, multiply this by given intensity:

  $\text{=}\left(\text{2}{\text{.83×10}}^{\text{23}}{\text{m}}^{\text{2}}\right)\left(\text{1350}{\text{W/m}}^{\text{2}}\right)\text{ =}\text{3}{\text{.82×10}}^{\text{26}}\text{W}$

Thirdly, to calculate an annual energy output, multiply by the number of seconds in a year:

  $\text{E}\text{=}\left(\text{3}{\text{.82×10}}^{\text{26}}\text{W}\right)\left(\text{3600}\frac{\text{s}}{\text{h}}\right)\left(\text{24}\frac{\text{h}}{\text{d}}\right)\left(\text{365}\text{.25}\frac{\text{d}}{\text{y}}\right)\text{=1}\text{.20}{\text{×10}}^{\text{34}}\text{J/y}$

Now, to calculate the number of photons per year, divide by the energy of one photon:

  $\frac{\text{E}}{\text{hf}}\text{=}\frac{\text{Eλ}}{\text{hc}}$

  $\frac{\left(\text{1}\text{.20}{\text{×10}}^{\text{34}}\text{J/y}\right)\left(\text{550}{\text{×10}}^{\text{-9}}\text{m}\right)}{\left(\text{6}\text{.63}{\text{×10}}^{\text{-34}}\text{J}\cdot \text{s}\right)\left(\text{3}\text{.00}\text{×}{\text{10}}^{\text{8}}\text{m/s}\right)}\text{= 3}\text{.33}\text{×}{\text{10}}^{\text{52}}\text{Photons/y}$

Conclusion:

The required value is

  $\text{3}\text{.33}\text{×}{\text{10}}^{\text{52}}\text{Photons/y}$ .