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Wien's Law 1) The sun has a temperature of 5800K. What is the sun's wavelength of maximum emission? Solution 2) The Earth has an average temperature of 288K. What is the Earth's wavelength of maximum emission? Solution
Stefan Boltzmann Law 1) How much more energy is emitted by an object that has a temperature of 100K compared to an object that has a temperature of 10K? Solution
Electromagnetic Wave Problems 1) A ray, emitted from the sun, is shining through your kitchen
window into a prism. The prism then casts a rainbow on the windowsill.
You just so happen to have a hand-held radiometer handy; closing
your eyes you place the instrument on a specific color of the rainbow.
You then open your eyes to see that the radiometer measured the
energy from that color at 3.0 x 10-19 joules. Given Plank's
constant of 6.6256 x 10-34 joules/sec, what possible
color did the radiometer measure? How do we know this? As stated before, light is measured in photons. These photons are measured in small units of length called nanometers (nm). Let's combine both equations and really test your comprehension. 2) Your teacher dropped a bag of Skittles on the floor: what a mess! She thought, "What a perfect problem for my intelligent students!" Picking up one of the tasty candies, she carefully measures the wavelength of one of the colors at 600 nm: boy is she ever smart! Her question is: how many joules of energy are contained in a photon with a wavelength of 600 nm?
SOLUTIONS 1) .5 microns HOW TO SOLVE PROBLEM: HOW TO SOLVE PROBLEM:
1) 10,000 times more energy HOW TO SOLVE PROBLEM: The amount of energy emitted by an object is directly related to the fourth power of the object's temperature. E = T^4, where E is the amount of energy emitted by an object per square meter per second and T is the temperature of the object in Kelvin. E = T4 E = (100)4 = (102)4 = 108 for the object with a temperature of 100K E= (10)4 = 104 for the object with a temperature of 10K Energy emitted by object with temperature of 100K / Energy emitted by object with temperature of 10K = 108/104 = 104 = 10,000 times more energy emitted by the object that has a temperature of 100K.
1) First we need to consider what would be the best equation to use. Obviously, the equation involving energy change, Plank's constant, and frequency is the best way to go. Next, we need to figure out what we are solving for. In this problem, they ask for the possible color that you measured. If we relate the energy with Plank's constant, we can solve for frequency. We are given the energy, 3.0 x 10-19 joules, as well as Plank's constant, 6.6256 x 10-34 joules/sec. Also, we are given the frequencies emitted by the visible spectrum, from red to violet. This problem is too easy now! If we solve for the frequency, we can then relate it to the energy emitted, measured in either sec-1, Hz, or less commonly, cycles/sec. Let's solve: -What is the best equation to use? -Solve for the intended variable. -What is given? Energy change = 3.0 x 10-19 joules
If the math is done correctly one should get 4.5 x 1014 sec1.
This falls within the given visible spectrum frequencies. Being
a low frequency on the visible spectrum, close to the frequencies
emitted by red colors, one could assume that you measured a color
close to red; though orange and yellow cannot be out of the question
as the given frequencies are a bit vague.
2) Again there are many conversions in
this problem. Let's attack it piece by piece. First we have an equation,
the speed of light, 3.0 x 1010 cm/sec, and the given
wavelength, 600 nm. We can relate these two parts once they have
units that can cancel. When they have similar units, we can solve
for one of the equations, C = ƒ * -What are we solving for in the end? -What is given? C = ƒ *
3.0 x 1010 cm/sec * 1 m/100cm = 3.0 x 108
m/sec
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and
E
= h *