Electromagnetic Radiation:

Problem 5.31:  Which has the higher frequency, infrared light or ultraviolet light?  Which has the longer wavelength?  Which has the greater energy?

The easiest way to do this one is to refer directly to a picture.  Here it is.

Just looking at the diagram the answer should be obvious unless you are a real dim bulb.  IR is seen to have the longer wavelength (l increases as we go left to right).  Noting that the frequency, n, increases as we go right to left, it should be immediately obvious that UV has the higher (larger ) frequency.  These things should be apparent from the following triptych of equations.  If you cannot comprehend this, then--in accordance to the ideas of radiation--you are a very dim bulb.

What about the energy?  The UV radiation clearly has the greater energy.  This is very clearly shown by the energy-frequency relationship,

That is the energy of radiation (in terms of its individual photons) is directly proportional to the frequency.  The equality is realized by introducing the physical constant h.
 

Two Constants it is Well to Know!

The speed of light	c = 2.99792458 x 108 m/s (exactly)
Planck's constant	h = 6.626069 x 10-34 J.s

Some numbers are best memorized to make your work faster (or, alternatively, you can store them in a pocket calculator or tatoo them on your forearm).  c will probably never change now that it has been defined exactly.  h changes a little from time to time as our methods of measurement improve. The values for various physical constants given here are the values which will be used in this set of problems.  Older values may be used elsewhere and, in the future, newer values may appear in even newer notes.  However, selections given here should affect answers hardly at all to the usual degree of precision that we have.
 

This is fairly easy.  All you need to is substitute into the equation shown here and solve:

The number of sig. figs. requested here is pathethic.  The authors seem to shy away from anything challenging.  Note that the speed of light is entered properly since that should be the number you put in your calculator.  Do not round until done!

Just rearrange the formula and plug in the numbers.  Note that 1 Hz = 1 s^-1.

The authors are trying to make things easy for you by not using metric prefixes and by not using Hz.  Don't worry.  We'll be sure that you learn these things by stating now that stuff you should have learned in the first three chapters will be in any exam handling the next few chapters!

Before going in to problems like these, it is valuable to get some conversion factors.  For instance, kJ/mol are commonly used in chemistry.  These (along with kcal/mol) are the most common way of expressing the energies of chemical reactions.  In the case of radiation, these are often how we express the energies of photochemical processes.

Since E = hn, the Hz is actually an energy unit.  In fact, a single Hz would have an energy of 6.626069 x 10^-34J.  In terms of kJ/mol, we can then get the following conversion factor:

To put this into another form, we could express this simply as

and now we have a convenient factor the the Hertz as an energy unit!  Now, let us do the problems.

These are both straight plugins (as long as you watch your frequency units)!  We do both at once now:

These numbers are considerably lower than typical energies for chemical reactions, as we shall see much later.  Also, these are much lower than energies associated with physical changes such as melting and boiling.  Obviously, the FM radio wave has higher energy than the AM radio wave.

X-rays, on the other hand, are somewhat above normal chemical reactions in energy.  This is easily seen with this example.  Note how we combine two equations, using simple algebra.  (If you find this difficult, you do not belong in this course.)  The x-ray energy comes out to be

X-rays correspond to energies needed to remove inner electrons from an atom and, on a kJ/mol basis, would have quite high energies as a consequence.

Microwaves, on the other hand, correspond to molecular rotational energies.  In the case of the microwave mentioned above, the total energy in kJ/mol would be

These energies are just under those of chemical reactions (which are a few orders of magnitude larger).

You'd have to have scrambled eggs for brains if you cannot see that x-rays have much more energy than microwaves!

We can use the same coversion factor as used in the same problem.  However, we now DIVIDE by it instead of multiplying!  The answers are in the following table and we show our work where necessary.

Here is our work for the problem above.  Note that we simply refer to the wavelength chart when needed to fill in the fourth column.  A copy of that is given here to save you some scrolling time.

The work:

The chart:

Problem 5.40:  What is the de Broglie wavelength (in meters) of a baseball weighing 145 g and traveling at 156 km/h?

Problems like this are simple but give "strange" answers.  For instance, the wavelength you get in this problem is far smaller than an atomic nucleus.  We shall do this now with all work shown.  As usual we start with the basic formula and then put in the correct number with conversion factors.

This is more or less the same type of problem--just the numbers are changed.  We not two things:

• The smaller the object (i.e., the less mass it has), the larger the wavelength.
• The slower the object, the longer the wavelength.

We now work the problem and observe our results.

The mosquito is smaller and slower and the wavelength, while still small, is much larger.

Nasty question:

Can any of you answer this?  Get on your web page discussion board and work it out among yourselves!

While this is not at a relevant wavelength, this does explain how electron diffraction and electron microscopes function.  With these, a beam of electrons at a high velocity can look at objects smaller than can be seen with an optical microscope.  For this problem, we show now that very small objects can have wavelengths which enter into the atomic and molecular size domains.  


Here is the answer without additional comment:

As you can see, this is a velocity not difficult to imagine!