Two Equations Governing Light's Behavior: Part Two
E = hν

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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 Two: E = hν

Brief historical note: It is well-known who first wrote this equation and when it happened. Max Planck is credited with the discovery of the "quantum," the discovery of which took place in December 1900. It was he who first wrote the equation above in his announcement of the discovery of the quantum.

1) E is the energy of the particular quantum of energy under study. When discussing electromagnetic quanta (of which light is only one example, x-rays and radio waves being two other examples), the word photon is used. A photon (the word is due to Albert Einstein) is a quantum of electromagnetic energy. The word quantum (quanta is the plural) is usually used in a more general sense, to describe various ideas of quantum theory or even, as I just did, to describe the entire theory itself.

2) h stands for a fundamental constant of nature now known as Planck's Constant. By the way, the discovery of the quantum had, and continues to have, many profound effects. Enough so that all of science (especially physics) before 1900 is refered to as "classical" and the science since 1900 is called "modern."

The value for Planck's Constant is 6.6260755 x 10¯34 Joule second. Please note that the unit is Joule MULTIPLIED BY second. It is not a division, both Joule and second are in the numerator.

3) ν is the frequency of the particular photon being studied. The discussion about frequency in part one applies here.

Before going on, I want to discuss one point. Frequency is a wave-based idea. What is it doing in a particle-based idea like the quantum? Good question. So much so that the term "wave packet" is often used in discussing these ideas. Indeed, modern science now speaks of "wave-particle duality" rather than "Light is a wave" or "Light is a particle." This whole area is profound and can lead to years of probing discussion. Albert Einstein and Niels Bohr (who were great friends) discussed these issues (and more) often over a period of many years, especially in the late 1920's and early 1930's. Their discussions are still important enough to merit historical study today. Every year, several books are published which delve into one or more of the implications of "wave-particle duality."

Last comment: you sometimes see the frequency identified using the letter f rather than the Greek letter nu. As in:

E= hf

Be aware of what it means, just in case you happen to see it.

Example #1:

(a) Identify λν = c as either a direct or inverse mathematical relationship.
(b) Do the same for E = hν.
(c) Write a mathematical equation for the relationship between energy and wavelength.
(d) Identify the equation for (c) as either direct or inverse.


(a) λν = c is an inverse relationship. As one value (say the wavelength) goes up, the other value (the frequency) must go down. Why? Because the product of the two must always equal the same value, c, which is a constant.

(b) E = hν is a direct relationship. As the frequency increases, so does E. Why? Because h remains constant. Notice that in this equation the two quantities which can change (ν and E) are on DIFFERENT sides of the equation whereas in the inverse relationship in (a), both λ and ν are on the same side, with the constant on the opposite side.

(c) Rearrange the light equation thusly:

ν = c / λ

Substitute for ν in E = hν:

E = hc / λ <--- use this equation when asked to calculate E after being given the wavelength

Rearrange to group the variables and the constants:

Eλ = hc

(d) The mathematical relationship is an inverse one. Note the equation's similarity to λν = c, with two values that can vary on the left side and a constant (h times c) on the right.

Example #2: How many Joules of energy are contained in a photon with λ = 550 nm? How many kJ/mol of energy is this?


1) Use ν = c / λ to get the frequency:

ν = (3.00 x 108 m s¯1) / (550 x 10¯9 m)

ν = 5.4545 x 10141.

I left a couple guard digits in the answer. Also, notice that the wavelength is not in scientific notation. This is because I made a silent conversion from nm to m. I didn't bother to convert it because it wasn't needed. be careful, your teacher may want you to have all numbers in scientific notation.

2) Now use E = hν to get the energy:

E = (6.626 x 10¯34 J s) (5.4545 x 10141)

E = 3.614 x 10¯19 J

This is the energy for one photon.

3) The last step is to find the kilojoules for one mole and for this we use Avogadro's Number:

x = (3.614 x 10¯19 J/photon) (6.022 x 1023 photon mol¯1) = 217635.08 J/mol

Dividing the answer by 1000 to make the change to kilojoules, we get 217.6 kJ/mol.

A slightly different way would be to use Eλ = hc (with the wavelength in meters) and solve for E, then multiply the answer times Avogadro's Number. Finally, divide by 1000 to get kJ/mol.

Example #3: How many kJ/mol (remember: mol means mole of photons) of energy is contained in light with a wavelength of 496.36 nm?


1) Convert 496.36 nm to meters:

496.36 nm times (1 m / 109 nm) = 496.36 x 10¯9 m

2) Use Eλ = hc

(E) (496.36 x 10¯9 m) = (6.626 x 10¯34 J s) (3.00 x 108 m s¯1)

E = 4.0048 x 10¯19 J

3) Multiply by Avogadro's Number:

4.0048 x 10¯19 J times 6.022 x 1023 mol¯1 = 241166 J/mol

241 kJ/mol (to three sig figs)

Example #4: What is the energy of a photon of green light with a frequency of 5.76 x 10141.


x = (6.626 x 10¯34 J s) (5.76 x 10141)

x = 3.82 x 10¯19 J

Comment: all frequencies of visible light will have an energy in the 10¯19 J range of values. If you wish to, you may calculate this for yourself. The wavelength range of visible light is taken to be from 400 nm to 700 nm. This translates (more-or-less) to a range from 5 x 10¯19 J down to 3 x 10¯19 J.

Example #5: When excited, some atoms produce an emssion with a frequency of 7.25 x 1012 Hz.

(a) calculate the energy, in Joules, for one photon with this frequency.
(b) calculate the energy, in kJ/mol.
(c) Is this light visible? Why or why not?

Comments on the solution:

(a) use E = hν

(b) use Avogadro's number as well as the answer from (a). Make sure to convert from the J value (which is what you'll calculate) to the kJ value.

(c) Calculate the wavelength using λν = c. What you need to do is compare the wavelength you calculate to the commonly accepted range of visible wavelengths, which is 400 nm to 700 nm. The wavelength you calculate will probably be in meters, so you will need to convert it to nm, then compare.

Example #6: Calculate the wavelength and frequency of a photon having the energy of 8.93 x 1010 J/mol.


1) Determine the energy of a single photon:

8.93 x 1010 J/mol / 6.022 x 1023 photon/mol = 1.4829 x 10¯13 J/photon

2) Determine the frequency of the photon:

E = hν

1.4829 x 10¯13 J = (6.626 x 10¯34 J s) (ν)

ν = 1.4829 x 10¯13 J / 6.626 x 10¯34 J s

ν = 2.238 x 1020 s ¯1

I'll leave it unrounded. If you were reporting this value to a teacher, you would report 2.24 x 1020 s ¯1

3) Determine the wavelength of the photon:

λν = c

(λ) (2.238 x 1020 s ¯1) = 3.00 x 108 m/s

λ = 3.00 x 108 m/s / 2.238 x 1020 s ¯1

λ = 1.34 x 10¯12 m

By the way, this is a very short wavelength. Visible light wavelengths are mostly in the 10¯7 m region.

4) Often, the wave length is asked for in nanometers:

(1.34 x 10¯12 m) (109 nm / 1 m) = 1.34 x 10¯3 nm

Or, λ = 0.00134 nm

5) Often, the region of the electromagnetic spectrum is asked for.

The best way to solve this is to search for images that show the various regions, from gamma to visible to radio waves. I searched and selected two (of many). Here is a link to them: EM Spectrum display page I also put the answer there.

I decided to put everything on a separate page so as to allow you to do your own searching to determine the answer, if you so desired.

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