This example problem demonstrates how to find the energy of a photon from its wavelength.To do this, you need to use the wave equation to relate wavelength to frequency and Planck's equation to find the energy. This type of problem is good practice at rearranging equations, using correct units, and tracking significant figures.

### Key Takeaways: Find Photon Energy From Wavelength

- The energy of a photo is related to its frequency and its wavelength. It is directly proportional to frequency and inversely proportional to wavelength.
- To find energy from wavelength, use the wave equation to get the frequency and then plug it into Planck's equation to solve for energy.
- This type of problem, while simple, is a good way to practice rearranging and combining equations (an essential skill in physics and chemistry).
- It's also important to report final values using the correct number of significant digits.

## Energy from Wavelength Problem - Laser Beam Energy

The red light from a helium-neon laser has a wavelength of 633 nm. What is the energy of one photon?

You need to use two equations to solve this problem:

The first is Planck's equation, which was proposed by Max Planck to describe how energy is transferred in quanta or packets. Planck's equation makes it possible to understand blackbody radiation and the photoelectric effect. The equation is:

**E = hν**

where

E = energy

h = Planck's constant = 6.626 x 10^{-34} J·s

ν = frequency

The second equation is the wave equation, which describes the speed of light in terms of wavelength and frequency. You use this equation to solve for frequency to plug into the first equation. The wave equation is:**c = λν**

where

c = speed of light = 3 x 10^{8} m/sec

λ = wavelength

ν = frequency

Rearrange the equation to solve for frequency:

ν = c/λ

Next, replace frequency in the first equation with c/λ to get a formula you can use:

E = hν**E = hc/λ**

In other words, the energy of a photo is directly proportional to its frequency and inversely proportional to its wavelength.

All that remains is to plug in the values and get the answer:

E = 6.626 x 10^{-34} J·s x 3 x 10^{8} m/sec/ (633 nm x 10^{-9} m/1 nm)

E = 1.988 x 10^{-25} J·m/6.33 x 10^{-7} m E = 3.14 x ^{-19} J**Answer:**

The energy of a single photon of red light from a helium-neon laser is 3.14 x ^{-19} J.

## Energy of One Mole of Photons

While the first example showed how to find the energy of a single photon, the same method may be used to find the energy of a mole of photons. Basically, what you do is find the energy of one photon and multiply it by Avogadro's number.

A light source emits radiation with a wavelength of 500.0 nm. Find the energy of one mole of photons of this radiation. Express the answer in units of kJ.

It's typical to need to perform a unit conversion on the wavelength value in order to get it to work in the equation. First, convert nm to m. Nano- is 10^{-9}, so all you need to do is move the decimal place over 9 spots or divide by 10^{9}.

500.0 nm = 500.0 x 10^{-9} m = 5.000 x 10^{-7} m

The last value is the wavelength expressed using scientific notation and the correct number of significant figures.

Remember how Planck's equation and the wave equation were combined to give:

**E = hc/λ**

E = (6.626 x 10^{-34} J·s)(3.000 x 10^{8} m/s) / (5.000 x 10^{-17} m)

E = 3.9756 x 10^{-19} J

However, this is the energy of a single photon. Multiply the value by Avogadro's number for the energy of a mole of photons:

energy of a mole of photons = (energy of a single photon) x (Avogadro's number)

energy of a mole of photons = (3.9756 x 10^{-19} J)(6.022 x 10^{23} mol^{-1}) [hint: multiply the decimal numbers and then subtract the denominator exponent from the numerator exponent to get the power of 10)

energy = 2.394 x 10^{5} J/mol

for one mole, the energy is 2.394 x 10^{5} J

Note how the value retains the correct number of significant figures. It still needs to be converted from J to kJ for the final answer:

energy = (2.394 x 10^{5} J)(1 kJ / 1000 J)

energy = 2.394 x 10^{2} kJ or 239.4 kJ

Remember, if you need to do additional unit conversions, watch your significant digits.

## Sources

- French, A.P., Taylor, E.F. (1978).
*An Introduction to Quantum Physics*. Van Nostrand Reinhold. London. ISBN 0-442-30770-5. - Griffiths, D.J. (1995).
*Introduction to Quantum Mechanics*. Prentice Hall. Upper Saddle River NJ. ISBN 0-13-124405-1. - Landsberg, P.T. (1978).
*Thermodynamics and Statistical Mechanics*. Oxford University Press. Oxford UK. ISBN 0-19-851142-6.