Problem #1: The Cassini spacecraft is currently on its way to Saturn and will arrive in 2004. Right now, it is approximately 85 million km from Earth. How long do radio signals from this spacecraft take to reach controllers here on Earth?
Solution: This one is pretty straightforward if you remember that radio waves are light waves. You'll soon learn that this is one of my pet peeves. Even though you hear sound from your radio, the radio waves which propagate from the radio station to the receiver inside your radio are a form of electromagnetic radiation; they are light waves with wavelength much longer than that of visible light waves. As light waves, they propagate at the speed of light, which is 3 x 108 m/s.
The distance between the spacecraft and Earth is 85 million km, or 85 x 106 km. We can (and should) convert this value to meters as follows:
Now, distance = rate x time, so time = distance/rate (divide both sides of the equation by the rate). We know the distance and the rate so,
Problem #2: a) How many green photons
(wavelength = 500 nm) does it take to equal the energy of one gamma ray
photon (wavelength = 10-2 nm)?
b) How many of these gamma ray photons would it take to equal
the the energy required to light a 100-watt light bulb for one second
(100 J)?
Solution -- Part a: Let's start with the energy of a photon:
where the constant listed above will yield an energy in Joules for an input wavelength measured in meters.
So, the energy of a photon with wavelength 500 nm, or 500 x 10-9 m, is
Likewise the energy of a photon with wavelength 10-2 nm, or 1.00 x 10-11 m, is
OK, so now we have the energy in photons of both wavelengths, and we want to know how many 500nm photons we would need to equal the energy in one gamma ray photon. Think about this for a minute, and I bet you'll see that the number of photons we need is equal to the ratio of the energies of the gamma ray and 500 nm photons. For example, if a gamma ray photon has ten times the energy of a 500 nm photon, then you'll need ten 500 nm photons to equal the energy of one gamma ray photon. If the energy ratio is 20, then you'll need 20 500nm photons, and so on.
We can express this idea mathematically by saying that the energy in N 500 nm photons is equal to the energy in one 10-2 nm photon. We don't yet know what N is but, since the energy in N 500 nm photons is equal to the energy in one 10-2 nm photon, we can set up the following equation:
and, if we divide both sides of the equation by the energy in one 500 nm photon,
which says that N is equal to the ratio of the energies of the two photons, which is precisely what we came up with based on our intuition above. So now we're in a position to answer the question:
You might note that this ratio of energies is equal to the ratio of the wavelengths of the two photons. That's not a coincidence at all. In fact, there's an even easier way to do this problem, if you set up the ratio at the very beginning. Consider the following definitions for the energies of the two photons:
where the subscripts denote that the quantity (energy and wavelength) pertains to the 500 nm or 10-2 nm photon. We can then construct the ratio of the energies directly,
and since the h and c appear on the top and bottom of the ratio, they divide out, leaving
Somewhere deep in your high school math background is the memory that 1/(1/x) = x (you can't fool me; I know it's there somewhere). Using this perhaps newly-remembered information, we can simply the above expression to yield:
and with a lot less calculator work, we get the answer that you'll need 50,000 green photons to equal the energy of just one gamma ray photon.
Solution -- Part b: Now we're asked how many gamma ray photons we'll need in order to supply 100 Joules of energy to a light bulb. From above, we know that one gamma ray photon contains 1.99 x 10-14 Joules; therefore N gamma ray photons will contain
We want to know what N is if the total energy is 100 Joules, so
and therefore
Problem #3: a)The public radio station in
Lewisburg broadcasts at a frequency of 100.1 MHz ("mega-hertz"). What
is the wavelength of these radio waves?
b)I'm driving in my car at 55 mph (24.6 m/s) directly toward
the radio station's broadcast tower. What is the frequency shift (due
to the Doppler Effect) in the signal my car radio receives? Is the
received frequency higher or lower than the broadcast frequency?
Solution -- Part a: This is just an application of the speed-wavelength-frequency relation for any type of wave:
We are given the frequency of the radio waves in MHz. 1 MHz = 106, so the frequency is
In addition, since radio waves are just another form of light waves, we know that the speed of these waves is the speed of light, 3.00 x 108 m/s. So,
or, rearranging the equation,
but what kind of a unit is m/s/Hz? Well, recall that the unit of frequency, Hz, is just "per second" or 1/s. So the 1/s in m/s and the 1/s from Hz cancel, leaving units of m. Does this unit make sense for our answer? Sure! We're calculating a wavelength, which is a length and should have units of length. Meter are units of length, so these units are appropriate, and our answer is:
Solution -- Part b: This part of the problem is a straightforward application of the Doppler Effect, much like the problem we did at the end of the first lab. The Doppler formula is:
We already know some of these quantities. The rest frequency is the frequency emitted by the radio station, 100.1 MHz, or 1.001 x 108 Hz. The speed of the wave is the speed of light, since radio waves are light waves (have I mentioned that before?). In the problem, you're told that I'm driving in my car toward the radio station at a speed of 24.6 m/s, so that's the difference between the speed of the emitter (the radio station isn't moving) and the speed of the receiver in my car. Therefore, we have three out of the four variables in the Doppler relation, so we can solve for the fourth, which is the shift in the frequency received. So,
Multiplying both sides by 1.001 x 108 Hz, we get
Note that the units of the answer are Hz because the m/s on the top and bottom of the right hand side of the equation cancel out. These units make sense since we're looking for a shift in frequency, which should be measured in units of frequency, or Hz.
Lastly, we need to decide whether the received frequency is shifted to a frequency higher or lower than that emitted by the radio station. Since I'm toward from the radio station, the separation between successive waves is a little smaller than if I weren't moving. That means that the received wavelength is a little shorter, and therefore that the received frequency is little higher.
Incidentally, 8.2 Hz is a tiny change in frequency -- much smaller that the ability of most radios to tune. The frequency shift introduced by a car's motion toward or away from a radio station is so small that you don't have to retune your radio every time you speed up or slow down.