Atomic Spectra:

Problem 5.44:  According to the equation for the Balmer line spectrum of hydrogen, a value of n = 3 gives a red spectral line at 656.3 nm, a value of n = 4 gives a green line at 486.1 nm, and a value of n = 5 gives a blue line at 434.0 nm.  Calculate the energy (in kilojoules per mole) of the radiation corresponding to each of these spectral lines.

Sometimes, it is useful in problems such as these to set up everything we need in advance.  In the previous section, we obtained a conversion factor between kJ/mol and Hz.  Here, we can do much the same.  But, in this case, we shall proceed a little bit differently and work directly with wavelengths.  Watch the units carefully here!

First, let us set up a conversion factor, E_nm, which tells us how much energy there is in radiation with a wavelength of exactly 1 nm.  If the answer is to be in kJ/mol, then we would do it something like this (again, I drop no significant figures because of the "Do not round until done" rule).

Now, we can use this same number in the three different calculations to get the answers directly.  In this case, we just use this shortcut:

It is quite often a powerful tool to combine the common parts of problems.  This cuts down on the number of calculations and makes use of the power of your calculator.  Also, we now have a useful number for energies for any wave whose wavelength is expressed in nanometers.  This is particularly important when we are being meticulous with our significant figures and our rounding rules.

Incidentally, if you are in to storing numbers in your calculators or computers, the following number for the number of kJ/mol for radiation with a wavelength of exactly 1 m might be more useful:

Now, you can enter in the wavelengths directly.  For instance, radiation of 35.0 mm wavelength could be entered directly as 35.0 x 10^-3 m and you would get an energy of 3.418 x 10^-3 kJ/mol.  Try this with other wavelengths to see what happens.

where m = 4, R = 1.097 x 10^-2 nm^-1, and n is an integer greater than 4.  Calculate the wavelengths (in nanometers) and energies (in kilojoules per mole) of the first two lines in the Brackett series.  In what region of the electromagnetic spectrum do they fall?

The Rydberg constant here is rounded off.  To make things more fun, we shall use the correct value (which will be shown in the solution below).  As is usual with me, I shall show how to do this in terms of formulas.  We shall then get the answers from simple plug-ins.  Note that we shall make use of the conversion factor obtained in the previous problem (5.44) we did.  (If you do something once and do it right, why do it again?)

Now, some algebra:

The little trick at the end of the first line gives a formula a little bit easier to do numerically.  The second formula just above is the wavelength in nm.  The little bit of skullduggery in the definition at the end of the second formula is just an example of how we often notate our variables to that we can remember what they are.  The energy of the given line is then simply

We are now ready to give our answers.  We shall do so in the following elegant little table.  It is left as an exercise to do the actual calculations and to be sure that your instructor did not goof!  (The first two lines in the Brackett series would be for n = 5 and 6.)
 

Ye Grande Analysis of Energies for Ye Series of Brackett:
Did Johnston Goof?

n	l (nm)	E (kJ/mol)
5	4050.0764	29.53687
6	2624.4495	48.48159

As you can see, the methods we stress in the web pages are aimed toward getting quick, direct, accurate, and precise answers.  And, it is a pretty good guess that Johnston did not goof here!

This is a simple plugin.  However, it does point out one important fact:

The energies we measure in spectroscopy are usually differences between energy levels and not energy levels themselves!

Now, we just do the problem (no fanfares or other decorations here):

Note that I did it two ways:  in nanometers and in meters directly.  Both give the same answer, of course!