Rydberg Formula

Q: Why does the formula use one over the wavelength?

A photon's energy is proportional to $1/\lambda$, so the energy drop between levels becomes a simple difference of fractions. Taking one over the wavelength gives a number of waves per meter, which makes the pattern easy to write.

In one sentence: The Rydberg formula gives the exact color of the light that a hydrogen atom gives off when its electron drops from one energy level to a lower one.

$\frac{1}{\lambda} = R_H\left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right)$

Read it out loud: “One over lambda equals R sub H, times, in brackets, one over n sub 1 squared, minus one over n sub 2 squared. Lambda (LAM-duh) is the wavelength of the light. R sub H is the Rydberg constant for hydrogen. n sub 1 and n sub 2 are two energy levels, whole numbers such as 1, 2 and 3, with n sub 2 the larger one. The small 2 after a number means squared, which is the number times itself.”

Before you start

The ideas and signs this page uses, explained from zero. For powers of ten, Greek letters and other notation in general, see Math basics.

Powers of ten, like $10^{7}$ and $10^{-9}$: The Rydberg constant is $1.09678 \times 10^{7}$ per meter. The $10^{7}$ means 1 followed by seven zeros, which is 10,000,000, so the constant is about ten million. A nanometer is $1\ \text{nm} = 10^{-9}$ m. The minus sign means divide by ten nine times, so a nanometer is one billionth of a meter. To multiply powers of ten, add the exponents: $10^{7} \times 10^{-9} = 10^{-2}$ .
Wavelength $\lambda$ (lambda) and the nanometer: The Greek letter $\lambda$ , said LAM-duh, is the usual name for wavelength, the distance from one wave crest to the next. For light, wavelengths are often given in nanometers. Visible light runs from about 400 nm for violet to about 700 nm for red. A wavelength near 656 nm is red, so the light from hydrogen is a color you can see.
Squares and fractions, like $\frac{1}{n^2}$: A square means a number times itself, so $n^2$ with $n = 2$ is $2 \times 2 = 4$ . The fraction $\frac{1}{n^2}$ is one divided by that square. For $n = 2$ it is $\frac{1}{4} = 0.25$ , and for $n = 3$ it is $\frac{1}{9} \approx 0.111$ . The bigger the level number, the smaller this fraction gets. Subtracting two fractions works like subtracting any numbers, once they share the same bottom number.
Units per meter, like $\text{m}^{-1}$: The unit $\text{m}^{-1}$ is read ‘per meter’, which means one divided by meters. A quantity in per meter counts waves in each meter. For example, $1/\lambda = 5$ per meter means one wave crest every 0.2 meters, because $1 \div 5 = 0.2$ . The Rydberg constant is in per meter, so the left side $1/\lambda$ is in per meter too.
Energy levels and light from an atom: An electron in an atom can only sit at certain energy levels, like rungs on a ladder, and it cannot sit between them. When the electron drops from a higher rung to a lower rung, the atom gives off the energy difference as one packet of light, called a photon. The photon’s color depends on its energy, so the size of the drop sets the color and the wavelength.
Subscripts, like $n_1$ and $n_2$: A small number written below a letter is a subscript. It tells similar quantities apart. $n_1$ and $n_2$ are two different level numbers, read ‘n one’ and ‘n two’. They are not multiplied together. The subscript H in $R_H$ means the constant belongs to hydrogen. Subscripts are only labels, and they never change the value of a letter.

What each symbol means

Symbol	Meaning
$\lambda$	The wavelength of the light the atom gives off or absorbs, in meters (m) inside the formula. Visible hydrogen light has wavelengths of about 400 to 700 nanometers, which is $4 \times 10^{-7}$ to $7 \times 10^{-7}$ m. Convert nanometers to meters before you use the formula.
$R_H$	The Rydberg constant for hydrogen, a fixed number that sets the scale of the whole pattern: $1.09678 \times 10^{7}$ m $^{-1}$ , about ten million waves per meter. For a nucleus of infinite mass the value is $1.0973731568 \times 10^{7}$ m $^{-1}$ . The small difference comes from the proton’s finite mass.
$n_1$	The lower energy level, the level the electron ends up on. It is a whole number 1, 2, 3 and so on, and it must be smaller than $n_2$ . Level 1 is the lowest, closest to the nucleus. The number has no unit, because it only counts levels.
$n_2$	The higher energy level, the level the electron starts from. It is a whole number bigger than $n_1$ . For example, if the electron falls from level 3 to level 2, then $n_2 = 3$ and $n_1 = 2$ . Like $n_1$ , it has no unit.

The formula piece by piece

$n_2$: Start with the higher level $n_2$ . It is the level the electron starts from, a whole number such as 3. A bigger $n_2$ means the electron starts higher up, so it has more energy to give away when it falls.
$n_1$: The lower level $n_1$ , where the electron ends up, is also a whole number, and it must be smaller than $n_2$ . The smaller $n_1$ is, the bigger the energy drop, so $n_1 = 1$ gives the most energetic light of the pattern.
$\frac{1}{n_1^2} - \frac{1}{n_2^2}$: Take one over each level squared, then subtract. The result is a positive fraction that depends only on the two levels. For $n_1 = 2$ and $n_2 = 3$ it equals $\frac{1}{4} - \frac{1}{9} = \frac{5}{36}$ . It is positive because $n_2$ is bigger, so $\frac{1}{n_2^2}$ is smaller than $\frac{1}{n_1^2}$ .
$R_H\left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right)$: Multiply that fraction by the Rydberg constant $R_H$ . The constant sets the scale, and the answer is $1/\lambda$ , a number of waves per meter. This step turns a pure fraction into a size that belongs to real light.
$\frac{1}{\lambda} = R_H\left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right)$: The whole formula in one sentence: one over the wavelength equals the Rydberg constant times the difference between one over the lower level squared and one over the higher level squared, so the wavelength is the reciprocal of that result.

What the formula is doing, and why

An electron in a hydrogen atom can only sit at certain energy levels, numbered by whole numbers. When it falls from a higher level $n_2$ to a lower level $n_1$ , it gives off one photon of light. The Rydberg formula gives the wavelength $\lambda$ of that photon. What goes in is two whole numbers, the two levels. What comes out is the wavelength, in meters or nanometers.

Why one over the wavelength? A photon’s energy is linked to its wavelength by $E = hc/\lambda$ , so the energy is proportional to $1/\lambda$ . Writing the formula with $1/\lambda$ makes the energy drop between two levels a simple difference of two fractions. A large value of $1/\lambda$ means a big energy drop and a short wavelength, such as violet or ultraviolet light.

Why the squares? In hydrogen, the energy of level $n$ is a fixed amount divided by $n^2$ . The levels crowd together as $n$ grows. The gap between levels 1 and 2 is about 10.2 eV, but the gap between levels 5 and 6 is only about 0.17 eV. The bracket is the difference of two such fractions, so it gives the size of the energy drop.

Why the minus sign? The bracket is a subtraction: one over $n_1^2$ minus one over $n_2^2$ . Because $n_2$ is the larger level, its fraction is the smaller one, so the bracket comes out positive. A photon must have a positive energy, so $n_2$ must be the larger level. If you put the larger number in the $n_1$ place, the bracket turns negative and the answer makes no physical sense.

Where it applies. Johannes Rydberg built this general pattern from measured spectral lines. The formula is very accurate for hydrogen when you use the constant $R_H$ above. It also works for ions that have only one electron, such as singly ionized helium, if you multiply the constant by $Z^2$ , the square of the nuclear charge. Jumps that end on level 2 form the Balmer series, mostly visible light. Jumps that end on level 1 form the Lyman series in the ultraviolet. Atoms with several electrons have extra effects that shift their lines, so the simple formula is only a rough guide for them.

An everyday picture

Think of a staircase with uneven steps. Each step has a fixed height, and a jump from one step to another releases a fixed amount of energy. The Rydberg formula is a table of those step heights. From the table you can work out the color of the light that each jump gives off. Big jumps release a lot of energy and give short wavelengths, while small jumps give longer wavelengths.

Worked example: The H-alpha line: an electron falls from level 3 to level 2

A hydrogen atom has its electron on level $n_2 = 3$ , and it falls to level $n_1 = 2$ . We want the wavelength of the light it gives off, using $R_H = 1.09678 \times 10^{7}$ m $^{-1}$ . Then we will find the energy of the photon in electron volts (eV), a small unit of energy used for atoms and light.

Write down the two levels. The electron starts on level 3, so $n_2 = 3$ , and it ends on level 2, so $n_1 = 2$ . Level 3 is the higher one, as the formula requires, because $n_2$ must be larger than $n_1$ .
Work out the bracket. The bracket is $\frac{1}{n_1^2} - \frac{1}{n_2^2} = \frac{1}{2^2} - \frac{1}{3^2} = \frac{1}{4} - \frac{1}{9}$ . Using 36 as the shared bottom number gives $\frac{9}{36} - \frac{4}{36} = \frac{5}{36} \approx 0.13889$ . The bracket is a plain number with no unit.
Multiply by $R_H$ and tidy the power of ten. Multiply $1.09678 \times 0.13889 \approx 0.15233$ , and keep the $10^{7}$ with it: $0.15233 \times 10^{7}$ . Move the decimal point one place to the right, which makes the front number ten times bigger and lowers the exponent by one: $0.15233 \times 10^{7} = 1.5233 \times 10^{6}$ per meter. This is $1/\lambda$ .
Invert to find the wavelength. To get $\lambda$ , divide 1 by $1/\lambda$ , so $\lambda = \frac{1}{1.5233 \times 10^{6}}$ m. The reciprocal of 1.5233 is about 0.6565, and the reciprocal of $10^{6}$ is $10^{-6}$ . So $\lambda \approx 0.6565 \times 10^{-6}$ m, and moving the decimal point one place to the right gives $6.565 \times 10^{-7}$ m.
Convert meters to nanometers. One nanometer is $10^{-9}$ m, so a length in meters divided by $10^{-9}$ gives nanometers. Dividing $6.565 \times 10^{-7}$ m by $10^{-9}$ m gives $6.565 \times 10^{2} = 656.5$ nm. The exponents subtract: $-7 - (-9) = 2$ .
Check the color. Visible light runs from about 400 nm for violet to about 700 nm for red, so 656.5 nm sits in the red part of the rainbow. This red line is called H-alpha. It is one reason hydrogen gas glows reddish-pink in discharge tubes.
Find the photon energy in electron volts. The energy of a photon is $E = hc/\lambda$ . Using the shortcut $hc = 1239.84$ eV nm, which combines Planck’s constant and the speed of light, gives $E = 1239.84 / 656.5 \approx 1.889$ eV. One electron volt is $1.602 \times 10^{-19}$ J, so this photon carries about 1.89 eV.

Answer: The electron gives off light with a wavelength of about 656.5 nm, which is red. The photon carries about 1.89 eV of energy. The Rydberg formula gives the color, and the energy comes from the same drop between the two levels.

The numbers in this example are recomputed by script every time the site is built.

Common mistakes

Swapping the levels. $n_2$ must be the higher level. Putting the larger number in the $n_1$ place makes the bracket negative, which gives a meaningless wavelength.
Stopping at the bracket. The bracket is only part of the answer. Multiply by $R_H$ to get $1/\lambda$ , and then take one over it to get $\lambda$ .
Forgetting that the left side is $1/\lambda$ , not $\lambda$ . Many readers take the result as $\lambda$ directly, which gives the reciprocal of the right answer.
Using a wavelength in nanometers inside the formula. $R_H$ is in per meter, so the answer comes out in meters. To convert to nanometers, multiply by $10^{9}$ .

Where you will meet it

Identifying hydrogen in stars and glowing gas clouds from the Balmer lines in their light.
Spectroscopy labs, where students calculate line wavelengths and compare them with a measured spectrum.
Predicting the lines of other one-electron ions, such as singly ionized helium, by multiplying the constant by $Z^2$ .
Calibrating spectrometers, because the hydrogen lines are sharp, well-known reference wavelengths.

Questions people ask

What are $n_1$ and $n_2$?

They are the two energy levels in one jump. $n_2$ is the higher level the electron starts from, and $n_1$ is the lower level it falls to. Both are whole numbers, and $n_2$ is always larger.

Why does the formula use one over the wavelength?

A photon’s energy is proportional to $1/\lambda$ , so the energy drop between levels becomes a simple difference of fractions. Taking one over the wavelength gives a number of waves per meter, which makes the pattern easy to write.

Does the same formula describe absorption?

Yes. An electron absorbs a photon of exactly the same energy when it jumps up, so the same wavelengths show up as dark lines in an absorption spectrum.

Hydrogen Energy Levels: An electron in a hydrogen atom can sit only on certain energy steps, and when it jumps between steps it gives off or takes in light of one exact color.
Hydrogen-Like Atom Energy: An atom or ion with only one electron has energy levels like hydrogen’s, but a larger nuclear charge pulls every level deeper, so the electron is held more tightly.
Planck-Einstein Relation (E = hf): E = hf says each packet of light carries energy in proportion to its frequency: higher frequency means more energy per packet, and brightness does not change it.