Two Equations Governing Light's Behavior: Part One
λν = c

Wavelength-Frequency Problems #1 - 10      Go to Part Two of Light Equations
Wavelength-Frequency Problems #11 - 20      Return to Electrons in Atoms menu

There are two equations concerning light that are usually taught in high school. Typically, both are taught without any derivation as to why they are the way they are. That is what I will do in the following.

Equation Number One: λν = c

Brief historical note: I am not sure who wrote this equation (or its equivalent) first. The wave theory of light has its origins in the late 1600's and was developed mathematically starting in the early 1800's. It was James Clerk Maxwell, in the 1860's, who first predicted that light was an electromagnetic wave and computed (rather than measured) its speed. By the way, the proof that light's speed was finite was published in 1676 and the first reliable measurements of the speed of light, ones that were very close to the modern value, took place in the late 1850's.

Each symbol in the equation is discussed below. Also, right before the examples, there is a mention of the two main types of problems teachers will ask using the equation. I encourage you to take a close look at that section.

1) λ is the Greek letter lambda and it stands for the wavelength of light. Wavelength is defined as the distance between two successive crests of a wave. When studying light, the most common units used for wavelength are: meter, centimeter, nanometer, and Ångström. Even though the offical unit used by SI is the meter, you will see explanations and problems which use the other three. Less often, you will see other units used; picometer is the most common one among the less-often used wavelength units. Ångström is a non-SI unit commonly included in discussions of SI units because of its wide usage.

Keep in mind these definitions:

one centimeter equals 10¯2 meter
one nanometer equals 10¯9 meter
one Ångström equals 10¯8 centimeter

The symbol for the Ångström is Å.

Most certainly, you will need to move easily from one unit to the other. For example, notice that 1 Å = 10¯10 meter. This means that 10 Å = 1 nm. So, if you are given an Ångström value for wavelength and a nanometer value is required, divide the Ångström value by 10. If you can't make easy transitions between various metric units, you'd better go back to study and practice that area some more.

2) ν is the Greek letter nu. It is NOT the letter v, it is the Greek letter nu. It stands for the frequency of the light wave. Frequency is defined as the number of wave cycles passing a fixed reference point in one second. When studying light, the unit for frequency is called the Hertz (its symbol is Hz). One Hertz is when one complete cycle passes the fixed point in one second, so a million Hz is when a million cycles pass the fixed point in one second.

There is an important point to make about the unit on Hz. It is NOT commonly written as cycles per second (or cycles/sec), but only as sec¯1 (more correctly, it should be written as s¯1; you need to know both ways). The "cycles" part is deleted, although you may see an occasional problem which uses it.

A brief mention of cycle: imagine a wave, frozen in time and space, where a wave crest is exactly lined up with our fixed reference point. Now, allow the wave to move until the following crest is exactly lined up with the reference point, then freeze the wave in place. This is one cycle of the wave and if all that took place in one second, then the frequencey of the wave is 1 Hz.

In any event, the only scientifically useful part of the unit is the denominator and so "per second" (remember, usually as s¯1) is what is used. The numerator "cycles" is not needed and so its presence is simply understood and, if writing a fraction is necessary, a one would be used, as in 1/sec.

3) c (lower case) is the symbol for the speed of light, the speed at which all electromagnetic radiation moves when in a perfect vacuum. (Light travels slower when passing through objects such as water, but it never travels faster than when in a perfect vacuum.)

Both ways shown below are used to write the value. You need to be aware of both:

3.00 x 108 m/s
3.00 x 1010 cm/s

The actual value is just slightly less, but the above values are the one generally used in introductory classes. (sometimes you'll see 2.9979 rather than 3.00.) Be careful about using the exponent and unit combination. Meters are longer than centimeters, so there are less of them used above.

Since there are two variables (λ and ν), we can have two types of calculations:

(a) given the wavelength, calculate the frequency; use this equation: ν = c / λ

(b) given the frequency, calculate the wavelength; use this equation: λ = c / ν

One last comment: you sometimes see the letter f used for frequency, replacing the Greek letter nu. Like this:

c = λf

Most likely, it will not cause you problems, but I did want to mention it, anyway.

An interesting little light trivia: light travels about one foot every nanosecond. You might try and work out the proper calculation before checking the answer.

Example #1: What is the frequency of electromagnetic radiation having a wavelength of 210.0 nm?


1) Convert nm to m:

210.0 nm x (1 m / 109 nm) = 210.0 x 10-9 m

We can leave it right there or convert it to scientific notation:

2.100 x 10-7 m

Either way works fine in the following calculation. Check with your teacher to see if they have a preference. Then, follow their preference.

2) Use λν = c

(2.100 x 10-7 m) (ν) = 3.00 x 108 m/s

ν = 3.00 x 108 m/s divided by 2.100 x 10-7 m

ν = 1.428 x 1015 s-1

Example #2: What is the frequency of violet light having a wavelength of 4000 Å?

The solution below depends on converting Å into cm. This means you must remember that the conversion is 1 Å = 10¯8 cm. The solution:

λν = c

(4000 x 10¯8 cm) (ν) = 3.00 x 1010 cm/s <--- note use of cm/s rather than m/s

ν = 7.50 x 10141

Notice how I did not bother to convert 4000 x 10¯8 into scientific notation. If I had done so, the value would have been 4.000 x 10¯5. Note also that I effectively consider 4000 to be 4 significant figures.

Comment: be aware that the range of 4000 to 7000 Å is taken to be the range of visible light. Notice how the frequencies stay within more-or-less the middle area of 1014, ranging from 4.29 to 7.50, but always being 1014. If you are faced with this calculation and you know the wavelength is a visible one (say 5550 Å, which is also 555 nm), then you know the exponent on the frequency MUST be 1014. If it isn't, then YOU (not the teacher) have made a mistake.

Example #3: What is the frequency of EMR having a wavelength of 555 nm? (EMR is an abbreviation for electromagnetic radiation.)

1) Let us convert nm into meters. Since one meter contains 109 nm, we have the following conversion:

555 nm x (1 m / 109 nm)

555 x 10¯9 m = 5.55 x 10¯7 m

2) Inserting into λν = c, gives:

(5.55 x 10¯7 m) (x) = 3.00 x 108 m s¯1

x = 5.40 x 10141

Example #4: What is the wavelength (in nm) of EMR with a frequency of 4.95 x 10141?

1) Substitute into λν = c, as follows:

(x) (4.95 x 10141) = 3.00 x 108 m s¯1

x = 6.06 x 10¯7 m

2) Now, we convert meters to nanometers:

6.06 x 10¯7 m x (109 nm / 1 m) = 606 nm

Example #5: What is the wavelength (in both cm and Å) of light with a frequency of 6.75 x 1014 Hz?

The fact that cm is asked for in the problem allows us to use the cm/s value for the speed of light:

(x) (6.75 x 10141) = 3.00 x 1010 cm s¯1

x = 4.44 x 10¯5 cm

Next, we convert to Å:

(4.44 x 10¯5 cm) x (108 Å / 1 cm) = 4440 Å

I could also have used (1 Å / 10-8 cm) for the conversion. I have a practice of putting the one with the larger unit (the cm in this case) and then figuring out how many of the smaller unit (the Å) there are in one of the larger unit.

Example #6: Which of the following represents the shortest wavelength?

(a) 6.3 x 10¯5 cm
(b) 7350 nm
(c) 3.5 x 10¯6 m


1) Convert the wavelengths such that they are all the same unit. I choose to convert to nanometers and will start with (a):

(6.3 x 10¯5 cm) (109 nm / 102 cm) = 630 nm

One immediate conclusion is that (b) is not the correct answer.

2) The conversion for (c)

(3.5 x 10¯6 m) (109 nm / 1 m) = 3600 nm

(a) is the correct answer.

Wavelength-Frequency Problems #1 - 10      Go to Part Two of Light Equations
Wavelength-Frequency Problems #11 - 20      Return to Electrons in Atoms menu