Question Video: Finding the Energy of a Photon Given Its Wavelength | Nagwa Question Video: Finding the Energy of a Photon Given Its Wavelength | Nagwa

# Question Video: Finding the Energy of a Photon Given Its Wavelength Physics • Third Year of Secondary School

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What is the energy of a photon that has a wavelength of 400 nm? Use 6.63 × 10⁻³⁴ J⋅s for the value of the Planck constant and 3.00 × 10⁸ m/s for the value of the speed of light in free space. Give your answer in scientific notation to two decimal places.

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### Video Transcript

What is the energy of a photon that has a wavelength of 400 nanometers? Use 6.63 times 10 to the negative 34 joule-seconds for the value of the Planck constant and 3.00 times 10 to the eighth meters per second for the value of the speed of light in free space. Give your answer in scientific notation to two decimal places.

To answer this question, we will need to relate the energy of a photon to its wavelength. To help us do this, we are given a value for the Planck constant and also a value for the speed of light in free space. With this information, the relationship that we need is that 𝐸 is equal to ℎ𝑐 divided by 𝜆, where 𝐸 is the energy of the photon, ℎ is the Planck constant, 𝑐 is the speed of light in free space, and 𝜆 is the wavelength of the photon. This formula is physically identical to expressing the energy as the Planck constant times the frequency of the photon, with these two formulas being related by the fact that the speed of light is equal to the frequency of a photon times its wavelength.

Anyway, before we substitute values into our formula, we need to modify the wavelength of the photon slightly. This is because the Planck constant with units of joule-seconds and the speed of light with units of meters per second are both expressed in terms of basic SI units. However, nanometers are not basic SI units. So we need to convert nanometers into meters. One nanometer is 10 to the negative nine meters. So the wavelength of our photon is 400 times 10 to the negative nine meters.

Alright, now all that’s left is to substitute this value and the values that we are given for the Planck constant and the speed of light. We have 6.63 times 10 to the negative 34 joule-seconds times 3.00 times 10 to the eighth meters per second divided by 400 times 10 to the negative nine meters. Let’s start by working out the units. In the numerator, we have seconds and also per second, but seconds per second is just one. And then we have meters in the numerator and meters in the denominator. But meters divided by meters is also one. So the overall units are just joules. This tells us we’re on the right track because joules are a unit of energy. When we evaluate the numerical portion of this quantity, we get 4.9725 times 10 to the negative 19, and then the units are joules. Rounding to two decimal places, our final answer is 4.97 times 10 to the negative 19 joules.

Now, when it comes to photons, it is often convenient to express their energies as electron volts. And 4.97 times 10 to the negative 19 joules is approximately 3.1 electron volts. The reason we mentioned this is because if we look at the units for the Planck constant and for the speed of light, we have joule-seconds and meters per second. The product of these joule-seconds times meters per second is joules times meters, which is an energy times a length. If we express energy in terms of electron volts and length in terms of nanometers, it turns out that the Planck constant times the speed of light is almost exactly 1240 electron volt nanometers.

This means that as long as we express the wavelength of the photon in nanometers, then the energy of the photon in electron volts is very nearly 1240 divided by the wavelength. In fact, 1240 divided by 400 is exactly equal to 3.1. And our true answer is less than a tenth of a percent different from 3.1 electron volts. Indeed, in general, approximating ℎ𝑐 as 1240 electron volt nanometers will give answers that are accurate to within a small fraction of one percent.

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