Contemporary Challenges: Spring-2026
HW Week 9 (SOLUTION): Due 27 Friday

1. Photons per second from laser S1 5523S (Q4B.4 from textbook) A typical laboratory helium-neon laser produces about 1 mW (0.001 J/s) of light at a wavelength of 633 nm. How many photons per second does this laser produce?

   The power of laser \(P\), is the energy per photon \(E_{ph}\) multiplied by the number of photons emitted per unit time, \(N_{ph}/ \Delta t\): \begin{equation} P = E_{ph}\frac{N_{ph}}{\Delta t}. \end{equation} We compute \(E_{ph}\) for our given wavelength \begin{align} E_{ph}&=\frac{hc}{\lambda}\\ &=\frac{(3.00\times10^8)(6.63\times10^{-34})}{(633\times10^{-9})}\text{ J}\notag\\ &= \frac{1.99\times10^{-25}}{6.33\times10^{-7}}\text{ J}\notag\\ &=3.14\times10^{-19}\text{ J}\notag \end{align} and solve for \(N_{ph}/ \Delta t\) \begin{align} \frac{N_{ph}}{\Delta t}&=\frac{P}{E_{ph}}\\ &=\frac{10^{-3}}{3.14\times10^{-19}}\text{ s}^{-1}\notag\\ &=0.32\times10^{16} \text{ s}^{-1}\notag\\ &=3.2\times10^{15} \text{ s}^{-1}\notag\\ \end{align} so we conclude that \(3.2\times10^{15}\) photons are emitted from the laser per second.

2. Seeing stars S1 5523S

   Our sun radiates light energy at a rate of \(P_{\text{sun}} = 4\times 10^{26}\text{ J/s}\). Assume that most of this energy is yellow light. Estimate how far away a star like the sun could be from Earth and still be seen by an unaided human eye. Assume the diameter of the pupil is 8 mm when dilated (looking at the night sky).

   Sense making: Most stars we see in the sky are hundreds of light years away. What does this say about the luminosity (total power of visible light) from most of the stars that we see?

   Reminder: a photoreceptor in the eye requires about 500 photons within a 30 ms time period to observe something.

   The light radiated by the sun is equally distributed around any spherical surface that is centered on the sun. The fraction of that light entering my eye is equal to the ratio between the area of my pupil and the area of a sphere enclosing the sun at my radius. Thus fraction of the energy entering my eye is \begin{align} \text{fraction of sun's light going into my eye} &= \frac{\text{area of my pupil}}{\text{area of sphere}} \\ &=\frac{\pi r^2}{4\pi R^2} \\ &=\frac{r^2}{4 R^2} \\ \end{align} Where \(r\) is the radius of my pupil (\(r\) = 4 mm) and \(R\) is the distance between me and the center of the sun. We will treat \(R\) as a variable and determine how big we can make \(R\) and still see the sun).

   Google says that yellow light has \(\lambda=580\text{ nm}\), so the energy of a yellow photon is \begin{align} E_{\text{photon}} &= \frac{hc}{\lambda} \\ &= \frac{[6.6\times 10^{-34}\text{ J s}][3\times 10^{8}\text{ m/s}]}{[580\times 10^{-9}\text{ m}]} \\ &= \frac{20\times 10^{-26}}{6\times 10^{-7}}\text{ J} \\ &= 3\times 10^{-19}\text{ J} \approx 2\text{ eV} \end{align}

   For a photoreceptor cell in my eye to "see" a distant version of our sun, I need a minimum number of photons per unit time: \begin{align} \text{minimum rate of photons per unit time} &= \frac{500}{[30\text{ ms}]} \\ \end{align}

   This corresponds to a minimum power, \begin{align} P_\text{min} &= \frac{500E_\text{photon}}{[30\text{ ms}]} \\ &= \frac{500 [3\times 10^{-19}\text{ J}]}{[30 \times 10^{-3}\text{ s}]} \\ &= 50\times 10^{-16}\text{ J/s} \end{align}

   I want to find the distance \(R\) at which light energy enters my eye at a rate equal to \(P_{\text{min}}\).

   \begin{align} P_\text{min} &= P_\text{sun}\frac{r^2}{4 R^2} \\ R &= \sqrt{\frac{r^2P_\text{sun}}{4P_\text{min}}} \\ R &= \frac{r}{2} \sqrt{\frac{P_\text{sun}}{P_\text{min}}} \\ &= 2 \times 10^{-3} \text{ m}\sqrt{\frac{4\times10^{26}\text{ J/s}}{0.5\times 10^{-14}\text{ J/s}}} \\ &= 2 \times 10^{-3} \text{ m} \sqrt{8 \times 10^{40 }}\\ &= 5.7 \times 10^{17} \text{ m} \\ \end{align}

   Now, converting this distance to light years \begin{align} \text{one light year} \approx 10^{16} \text{ m} \\ R = 57 \text{ light years} \end{align}

   In conclusion, if a star has luminosity equal to our sun, we can only see it with the unaided eye if \(R\) < 57 light years. If most of the stars we see with the unaided eye are hundreds of light years away, those stars must have significantly greater luminosity than our sun.

3. Light from an electron in a box S1 5523S
   

   Suppose an electron is trapped in a box whose length is \(L= 1.2 \text{ nm}\). This is a coarse-grained model for an electron in a small molecule like cyanine (see Example Q11.1 in the textbook, and the figure above). If we solve the Schrodinger equation for this coarse-grained model, the possible energy levels for this electron are \begin{align} E = \frac{h^2 n^2}{8 m L^2} \end{align} where \(m\) is the mass of the electron and \(n= 1,\ 2,\ 3,\ ...\)

   Draw a spectrum chart (like the righthand side of Figure Q11.2) illustrating all visible spectral lines (i.e. any wavelengths between infrared and ultraviolet) that can be emitted by a test tube full of cyanine molecules when every possible electronic transition is happening.

   Note: Due to the shape/symmetries of electron wavefunctions in a box, optical transitions between energy levels only happen when \(\Delta n = n_\text{initial}-n_\text{final}\), is an odd integer.

   

   Visible photons

   

   The energy levels fro the electron are \begin{align} E_n &= \frac{h^2 n^2}{8 m L^2} \qquad \qquad \text{ where } L = 1.12\text{ nm}\\ &= \left[\frac{(6.6\times10^{-34})^2}{8 (9\times10^{-31}) (1.2\times10^{-9})^2}\right] n^2\\ &\approx [0.4\times10^{-19}\text{ J}]n^2\\ &=[0.25\text{ eV}]n^2 \end{align} First I'll look at transitions with \(\Delta n = 1\), for example \(n=4\) to \(n=3\).

   

   Now I'll check transitions with \(\Delta n=3\), for example \(n=4\) to \(n=1\).

   

   If \(\Delta n=5\) the system would emit photons of 3.75 eV or greater (outside the visible spectrum).

   In conclusion, there are four energies of visible photons that this system could emit: 2 eV, 2.25 eV, 2.75 eV and 3 eV.

4. Efficiency of a solar cell S1 5523S
   

   Download the file extraterr_solar.csv, which is in comma-separated-variable (csv) format. Open the csv file in a spreadsheet program such as Excel or Google Sheets. The data is the spectral intensity with respect to wavelength, \(S_\lambda\), for the sunlight that is hitting a satellite above the earth. The first column is wavelength in units of nanometers. The second column is spectral intensity in units of W/(m\(^2\cdot\)nm).

   1. Use a spreadsheet to perform a simple numerical integration (Riemann sum) to find the total energy flux hitting the satellite. Explain your method using summation notation. Additionally, write down the formula you enter in the spreadsheet (e.g. =SUM(B1:B745)). Give your final answer in units of W/m\(^2\) and check that it is reasonable.

      The numerical integration is performing \(\sum_{i=1}^{N} I_{\lambda, i} \cdot \Delta \lambda\) where \(I_{\lambda, i}\) is the \(i^{\text { th }}\) value of spectral intensity and \(\Delta \lambda=5\) nm. In Excel I used the formula =5*SUM(B1:B745). Performing this sum yields a total energy flux of \(I=1348 \frac{\mathrm{W}}{\mathrm{m}^{2}}.\)

   2. Consider a narrow band of wavelengths, from 552.5 nm to 557.5 nm. (The bandwidth is 5 nm and the central wavelength is 555 nm). All the photons in this bandwidth have very similar energy, \(E_{\text{photon}} \approx\) (1240 nm\(\cdot\)eV)/(555 nm). How many photons per second per \(\text{m}^2\) are in this spectral band of sunlight? Explain your method using standard mathmeatical notation. Additionally, write down the formula that you entered into the spreadsheet.

      The spectral intensity at 555 \(\mathrm{nm}\) is 1.86\(\frac{\mathrm{W}}{\mathrm{nm\cdot m}^{2}}\) and the photon energy is 2.23 \(\mathrm{eV}\). Multiplying the spectral intensity by the 5 nm spectral width and dividing by the photon energy gives a photon flux of \(I_{\mathrm{ph}}=2.6 \times 10^{19} \frac{\text { photons }}{\mathrm{s} \cdot \mathrm{m}^{2}}\). In Excel I used the formula =5*B56/(1.6E-19*1240/A56)

      The calculation that you did for part b can now be applied to every row in your spreadsheet. You will need these numbers for part c.
      

   3. Silicon solar cells absorb photons if \(E_{\text{photon}}> 1.1\text{ eV}\). That is to say, \(E_{\text{photon}}\) must be greater than gap between occupied and unoccupied quantum energy levels in silicon. Use your spreadsheet to calculate how many photons per second per \(\text{m}^2\) have sufficient energy to be absorbed by a solar cell. Write down the formula that you entered into the spreadsheet.

      In my spreadsheet I created column C which lists the number of photons in each spectral bin. I summed the photon numbers in these spectral bins, starting at a photon energy of 1.1 eV and going up to the highest photon energy (around 4.3 eV). In Excel I used the formula =SUM(C1:C170). The result is \(3.38 \times 10^{21} \frac{\text { photons }}{\mathrm{s} \cdot \mathrm{m}^{2}}\).

   4. The electrical energy produced by a silicon solar cell cannot exceed
      \((1.1\text{ eV})\ \times\text{ (number of absorbed photons)}\).
      Calculate the maximum possible rate that electrical energy could be produced by a solar cell attached to this satellite per unit area. Give your answer in units of \(\text{W/m}^2\).

      The power per area generated by the silicon solar cell is the total photon flux above the 1.1 eV cutoff (the result from part d) multiplied by the energy that can be harvested from each photon which is \(1.1 \mathrm{eV}=1.76 \times 10^{-19} \mathrm{J}\). Performing this calculation yields a total power per area of \(596 \frac{\mathrm{W}}{\mathrm{m}^{2}}.\)

   5. Compare your answers to part a and part d. What is the maximum possible efficiency of the solar cell (i.e. the ratio of the electrical energy output to the total energy input)?

      The ratio of the power per area found in part e) to the intensity of sunlight found in part a) is 44\(\%\). The rest of the sun's energy is either not absorbed (\(24\%\)) or turned into heat inside the silicon (\(32\%\)).
      
      This homework question is a coarse-grained model that respects the basic operating principle of a solar cell. However, it is not the complete story. The true efficiency limit for a solar cell made from a single pn junction, operating with natural solar intensity is about 33%.

      To understand this fully, we should study the physics of pn junctions. Then we would know the max volage we could expect from the solar cell (it is slightly less than the bandgap voltage). We should also know how much current can be extracted without drastically diminishing the voltage. (The current must be slightly smaller than 1 electron per absorbed photon).

      Shockley & Queisser published such calculations in 1961. The paper is 10 pages long. On page 3 they present a zeroth-order estimate, which is consistent with this homework question: \begin{align} &\text{"ultimate efficiency"}\notag\\ &=x_g\left(\int_{x_g}^{\infty}\frac{x^2\ dx}{e^x - 1}\right)/\left(\int_0^{\infty}\frac{x^3\ dx}{e^x -1}\right) \end{align} where \(x_g = E_{gap}/(k_\text{B} T_{sun})\) and \(T_{sun} \approx 6000\text{ K}\). The integrands in this equation come from Planck's law. The maximum efficiency, \(44\%\), occurs at \(E_{gap} = 1.1\text{ eV}\). The next 7 pages of Shockley & Queisser's paper considers the details of a real pn junction, with the fundamental limitations of working at room temperature with a limited intensity of solar radiation. They arrived at a limit of 33%.