ECE 5/6358 HW 2A (100 pts)
First name, Last name

Student ID:

Han Q Le(c)

Part A

1. (50 pts) Basic things about light

1.1 Quantum

We discuss Planck’s light quantum, the smallest anount of energy that a light carries, as: E=h ν where h is Planck’s constant and ν is light frequency. (h= 6.62607015 ×(10^(-34)) Joule/s). Don’t Google it, it is WRONG on Google. Find the photon energy for the following wavelength:

λ={0.85,0.78, 0.65,0.532,0.405} (* all in micron  *)

The speed of light is 299.792458 μm/ps

hint

Do NOT execute the code given below and just leave it there. Write a sentence, make a table to explain to the general audience like your parents, or friends (not to me, please!) what it means.

c=299.792458 ; hJ=6.62607015*(10^(-34));
laserλ={0.85,0.78, 0.65,0.532,0.405} ; 
laserν=c*10^12/laserλ ;
laserQ=hJ* laserν ;

1.2 The unit of eV

A most useful unit of energy is eV= electron-Volt: it is the energy to move an electron across a potential of 1 V. What is one eV in the unit of Joule. (note the charge of an electron is: e=1.602176634 ×(10^(-19)) Coulomb, which is exactly defined so).

hint

The answer is in the question itself.

1.3 Photon energy in unit of eV

What is the energy of each photon in question 1.1 in unit of eV?

hint

Do NOT execute the code given below and just leave it there. Write a sentence, make a table to explain to the general audience like your parents, or friends (not to me, please!) what you calculate.

c=299.792458 ; hJ=6.62607015*(10^(-34));
laserλ={0.85,0.78, 0.65,0.532,0.405} ; 
laserν=c*10^12/laserλ ;
laserQ=hJ* laserν ;
elementarycharge=1.602176634*(10^(-19));
laserQ/elementarycharge;

1.4 Energetic photon

Red-green light and UV light, which is more dangerous to human? Why? Write a ≤ 200-word essay about the danger of UV photons to human, while longer wavelength photons do not pose such danger. Do not plagiarize.

hint:

Just search about DNA damages and other type of damage to the skins. Look at this article for your amusement: https://www.science20.com/cool-links/unilateral_dermatoheliosis_28_years_photoaging_due_sun_exposure-92122
The original article is in New England Journal of Medicine, which requires subscription.

1.5 Astronomical photon energy

Below is the plot of gamma ray intensity as function of photon energy from the center of our galaxy (Milkyway) obtained by Max Planck Institute for Astrophysics. (γ-ray is photon with very high energy.)

Using the formula: E=m (c^2), and the electron rest mass is 9.10938215×(10^(-31)) Kg, guess what nature of the peak at 511 keV (kilo eV) in the graph is. Calculate the 511-keV-line intensity (the peak if the graph) in unit of W/(m^2)

hint

It is from the electron-positron annihilation. Do NOT execute the code given below and just leave it there. Write a sentence what the whole thing means to explain to the general audience like your parents, or friends (not to me, please!) (such as this is the evidence that anti-matter exists in the Universe),

c=299792458. ;
me=9.1093837015*(10^(-31)) ;
eC=1.602176634*(10^(-19));
me (c^2) /eC ;

2. (50 pts) The Sun and the Earth

2.1 (10 pts) Solar radiation

Below shows the light spectrum emitted from the Sun and from the Earth (intrinsically from Earth, not from the Sun illumination, in other words, the light emitted from the Earth on its dark side). Calculate the photon energy (in unit of eV) at the wavelength where it is brightest (λ at peak radiative flux) for the sun and earth. (You can refer to your plot in 1.7).

2.1 Answer

Sun: 0.5 μm, 2.48 eV
Earth: 10 μm, 0.124 eV

2.2 (10 pts) Sun intensity on Earth

The radiance at the Sun surface (the power emitted per unit area) is ~63 MW/(m^2) (megaWatt/(m^2)). Assume the Sun radiation is uniform over 4 π radian, and there is no loss of power from the Sun to the Earth, calculate the solar radiation flux (sun light intensity in unit of W/(m^2)) at the earth right above its atmosphere. You should look up the quantities you need to calculate such as the Sun radius and the distance between the Sun and the Earth.

hint

Do NOT execute the code given below and just leave it there. Write a sentence so that the general audience like your parents, or friends (not to me, please!) understand.

SunR=6.95500*(10^8) ; (* sun radius *)
SEdistance=1.49597870700 *(10^11) ;
PE=63. * (10^6)*(((SunR/SEdistance))^2)  ;

2.3 (10 pts) Cosine irradiance law

Download this app to study the cosine irradiance law

Then, answer this: Suppose on a clear sky, the Sun irradiance on the roof of your house is 1000 W/(m^2). You have a solar panel with an active area of 1.5-(m^2) that is perpendicular to the Sun light at 12 noon. Then later in the afternoon, assuming that the Sun irradiance is the same (in general, it is not because of larger atmostpheric loss, but we assume the same just for this problem), but the incident angle of the Sun relative to your solar panel is 50 degrees. Assume your solar panel has 20% efficiency, how much power it can generate in each case above.

2.4 (20 pts) The 4 seasons

The reason the Earth has seasons is because its rotation axis tilts relative to the plane of its orbit.

Download this app

Enter the latitude of your home town or city where you were born and grew up (please state it, and google search its latitude), then use the graphs of the app to estimate the relative solar power at the highest time on the 4 seasonal dates in the graph above. (we will soon have Autumnal Equinox). The time for highest solar irradiance is supposedly 12 noon, but different cities and towns have different time reference (such as Daylight Savings time). We don’t care exact time, just the highest value of the day.