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Blackbody Spectrum & Lasers SIM Homework Answer Key
1) (6 pts total) The Sun provides most of the energy that makes life on earth possible. The Sun’s
surface reaches a temperature of about 5800 °C and emits sunlight which travels the 150,000,000
km to Earth at a speed of 3*10^8 m/sec in about 8 minutes.
a) (1.5 pts) Electromagnetic radiation is emitted when charged particles accelerate. Sunlight
is a blend of electromagnetic waves of different wavelengths.
i) Describe the distribution of electromagnetic radiation emitted from the sun (what
wavelengths are emitted? is it all visible light? is there more blue light than red light?).
Click here for the blackbody spectrum simulation
The radiation from the sun takes the form of a blackbody spectrum. The
peak intensity is in the visible light, and there is also a significant
contribution from ultraviolet (higher energy than visible light) and infrared
(lower energy than visible). Within the visible portion of the spectrum, the
peak intensity is in the blue, around 500 nm.
ii) Why you get a distribution of wavelengths instead of just one wavelength?
The electromagnetic radiation from the sun is produced by the motion of
charged particles at the sun’s surface. As the particles move around rapidly
and collide with each other, their acceleration causes them to radiate
electromagnetic waves. Temperature is defined by the average kinetic
energy of the particles, so the temperature of the sun (or any black body)
determines the peak wavelength of light that will be emitted. However, some
of the charged particles will be moving slower and some will be moving faster,
so the charged particles actually have a range of energies distributed
around the average kinetic energy. As a result, the light they emit will have a
range of energies (and wavelength). The particles moving about (vibrating)
faster will produce higher energy light (shorter wavelength) than the ones
moving about a bit slower.
iii) How does the temperature of the sun affect this distribution?
Higher temperature means that the charged particles will on average be
moving about (vibrating) faster. More light will be produced at all
wavelengths and these faster moving particles will produce higher energy
light (shorter wavelength) shifting the peak of the spectrum to higher
energy light (towards the blue).
b) (1 pt) It is the thermal motion of charged particles at the sun’s surface that produces the
electromagnetic radiation emitted by the sun.
i) (0.5 pts) To generate a green light at 550nm, at what frequency would a charged
particle have to be vibrating back and forth?
We know that the wavelength and frequency of light are related by the
Speed of light = wavelength x frequency.
c = (lambda) x f
So f = c / lambda = (3 x 10^8 m/s) / (550 x 10^-9 m) = 5.45 x 10^14 Hz.
To produce light at this frequency, a charged particle would have to vibrate
at the same frequency.
ii) (0.5 pts)Even in the most advanced circuits, we cannot oscillate electrons back and
forth at that rate through wires. But we can oscillate charges back and forth quickly
enough to broadcast TV using radio wave signals. At what frequency does that
electronics at the TV station need to have the charges oscillate back and forth on a TV
broadcast antenna to transmit a typical TV signal (say a radio wave transmission signal
with a wavelength of 1 meter)?
For a TV signal with 1 m wavelength, f = (3 x 10^8 m/s) / (1 m) = 3 x 10^8 Hz.
c) (2.5 pts) The sun provides the energy that powers life on our planet everyday (even the
energy stored in fossil fuels came from the sun). At noon on a typical mid-June day in
Boulder, each 1 cm2 of the surface is exposure to 20 mW (milliWatts) of visible radiation.
You are considering adding solar panels to your house and want to see whether they will
provide enough energy for your household needs.
i) (0.5 pts) If your solar panels covered an area of 1 meter by 5 meters, how much solar
power would be hitting the surface of the panels at noon on a clear day? (in Watts ….
check your answer with your fellow students!)
We know that 20 mW of power are hitting each 1 cm2 of area. The solar
panels are 1 meter by 5 meters, which is the same as 100 cm by 500 cm,
so Area of solar panels = length x width = 100 cm x 500 cm = 50,000 cm2,
Solar power hitting surface = area of solar panels x power/area = 50,000
cm2 x 20 mW/cm2 = 1,000,000 mW = 1,000 Watts
ii) (0.5 pts) Over the period of 1 hour, what is the total amount of energy of the visible
light that has hit the solar panels? (assume that the exposure is pretty steady for that hour
at 20 mW).
A watt is a measure of power, which is amount of energy per second. Since a
joule is the unit of energy, watts are joules per second.
Total energy hitting solar panels in 1 hour = power hitting panels x the
amount of time in seconds.
Total energy = 1,000 Joules/second x 3600 seconds = 3,600,000 Joules
iii) (1 pt) Red photons and violet photons have slightly different energies. If we take the
average energy of the visible light photon hitting the solar panels as that of a green
photon (550 nm), how many visible light photons hit the solar panels during that hour?
First we need to figure out how much energy 1 green photon has.
Green light has a frequency of 5.45 x 10^14 Hz. The energy in one photon
and frequency are related by:
Energy of one photon = h x f = h x c/(lambda), where h is Planck’s constant.
So Energy of one photon of green = (6.626 x 10^-34 Js) x (5.45 x 10^14 Hz)
= 3.6 x 10^-19 J
We know that each photon contributes 3.6 x 10^-19 J (a very small amount
of energy) and our solar panel has had 3,600,000 Joules of energy hit it, so
we should expect a very, very large number of photons to have hit the solar
panel. In equations:
# of photons that hit = total energy/energy per photon = 3,600,000 Joules /
(3.6 x 10^-19 J/photon) = 1 x 1025 photons
iv) (0.5 pts) If the solar panels were 15% efficient at converting this energy into electrical
energy, how much electrical energy would have been harnessed in 1 hour?
Energy harnessed = 0.15 x energy hitting solar panel = 0.15 x 3,600,000
Joules = 540,000 Joules
v) (1 pt) For how long could this amount of energy power a 60 W bulb (in hours)?
A 60 Watt bulb uses 60 Joules per second. Since we know that we have
540,000 Joules of energy total and the bulb use 60 Joules each second, we
can calculate how many seconds we can run the bulb for.
# of seconds of bulb operation = 540,000 Joules / (60 J/s) = 9000 seconds
# of hours of operation = 9000 seconds x 1 minute/60 seconds x 1 hour/60
minutes = 2.5 hours
d) (1 pt) As we stated above, in Boulder at noontime in the summer there is 20 mW of visible
light hitting each 1 cm2 of the surface. However, there is only about 0.022 mW (milliWatts)
of UV radiation between 290 nm and 320 nm (UV B) hitting each 1 cm2 of the surface.
(Only a small portion of the sunlight’s energy is at UV wavelengths.) Explain why it is that
0.022 mW of UV exposure can cause damage to our DNA and result in skin cancer, but
exposure to 20 mW of visible radiation leaves our DNA unharmed? Why doesn’t exposure to
so much more visible radiation harm our DNA?
Each photon of UV light has enough energy to break apart a DNA molecule,
whereas a photon of visible light does not. While some of the visible light will be
absorbed by our skin, its energy will just be turned into heat and will not do any
2) (4 pts) A laser involves the various processes by which light interacts with atoms’ (1)
absorption – the process by which the light is absorbed and the energy causes the atomic electron
to go to a higher energy level (2) spontaneous emission – the process in which the electron jumps
back down and spits out a photon of the corresponding energy as it does so, and finally, (3)
stimulated emission – the process where a photon hits an atom that is in a higher energy level
already and this causes the atom to spit out a photon that is identical to the one that hit the atom
so there are two identical photons. Not every photon that hits a ground state atom will be
absorbed, and not every photon that hits an excited atom will cause stimulated emission, but the
probability of the event occurring during a photon encounter is the same for both conditions.
That means that if a photon is going through a gas of atoms and it encounters single ground state
atom its chance of being absorbed by that atom is exactly the same as its chance of causing
stimulated emission if it had encountered a given single atom in an excited state. So, if there are
twice as many atoms in the ground state as in the excited state the photon is twice as likely to be
absorbed as it is to stimulate an atom to emit.
a) (1 pt) Explain why it would be impossible to achieve a population inversion and hence
impossible to make a laser work if your design used an atom where only two energy levels
were involved (the lowest or ground level and one higher level) and in your design you were
“pumping” atoms into the excited state using light at the exact wavelength where the photon
energy matched the energy separation between the lower and upper energy levels of the atom.
(Note that: This is the reason that in actual lasers that use light to pump the atoms into the
upper levels, three or four atomic energy levels are involved.)
As light is sent into the gas of atoms, some of the photons will excite atoms to
a higher energy level. As the fraction of the atoms that are excited increases,
so does the probability that a photon will interact with one of these atoms and
stimulate emission of another photon. By the time roughly half of the atoms are
in the excited state, an incoming photon is equally likely to excite an atom that
is in the ground state or stimulate emission of a photon by an excited atom. If
we consider the average effect of many photons, the number of atoms in the
excited state will remain essentially constant and it will be impossible to achieve
a population inversion where most of the atoms are excited.
b) (1 pt) Explain the function of the mirrors at each end of a laser. Why do you need them to
have a functioning laser? What would happen if one mirror was removed?
One of the special properties of a laser is that the light is collimated; that is,
all of the light travels parallel rather than spreading out. Some of the photons
emitted by a laser come from spontaneous emission, and those will travel in
random directions. Most will miss the mirrors and will not become part of the
laser beam. But those that travel parallel to the laser cavity will hit the mirrors
and bounce back and forth many times before exiting as part of the laser beam.
On these several passes though the laser cavity, the photons have many more
chances to interact with atoms and stimulate emission of additional photons,
which will be identical and will travel in the same direction, contributing to the
collimated laser beam.
c) (1 pt) If you set up all the parts for a laser and then just barely turn on the pump that puts
atoms into the excited state nothing very interesting happens. As you gradually increase the
pumping you still do not get any laser light out. However, once you reach a certain point
(called the “threshold”) suddenly laser light comes out and as you increase above that the
amount of laser light produced increases rapidly. This behavior can be observed in the PhET
laser applet as well as in any real laser. Explain this behavior of a threshold for laser
operation in terms of the underlying physics of lasers.
If the pump is on at low intensity, some atoms will be excited by photons but
most will emit a photon spontaneously, in a random direction, rather than
through stimulated emission. Since most of the atoms are in the ground state,
these photons are unlikely to encounter another excited atom, and will just pass
out of the laser medium in a random direction. However, if the intensity is high
enough, a population inversion is achieved, and any photon traveling though the
gas has a much higher probability of encountering an excited atom and
stimulating emission of another identical photon. Many of these photons will
bounce back and forth between the mirrors, stimulating emission of more and
more photons on their way and producing an intense beam of collimated light.
The threshold is the point where the pump light is intense enough to ensure
that most of the atoms are in the excited state, and free to emit a photon, at
any given time.
d) (1 pt) Lasers are used in all sorts of modern technology. What characteristics of laser light
make it special and especially useful in technology (try to give examples of where this
feature is made use of in your everyday life)? In your description of the characteristics,
explain why the physics of how lasers work produce these characteristics.
Laser beams are light of all the same wavelength (monochromatic) with all of
the light traveling in the same direction (well-collimated) and all of the light has
the same phase (the wave of light created by any one atom photon is exactly in
step with the rest of the light). These are all characteristics that are
associated with stimulated emission (spontaneous emission is not guaranteed to
do any of these things). Stimulated emission occurs when a photon, passing by
an excited atom, stimulates an atom to emit a photon of the same wavelength, in
the same direction, and of the same phase (in step with the other photon).