# Question Video: Understanding a Photon Emitted By an Electron Transitioning From Energy Level 3 to 2 Physics • 9th Grade

The diagram shows the transition of an electron in a hydrogen atom from 𝑛 = 3 to 𝑛 = 2, emitting a photon as it does so. What is the energy of the photon? What is the wavelength of the photon? Use a value of 4.14 × 10⁻¹⁵ eV⋅s for the value of the Planck constant. Give your answer to the nearest nanometer.

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### Video Transcript

The diagram shows the transition of an electron in a hydrogen atom from 𝑛 equals three to 𝑛 equals two, emitting a photon as it does so. What is the energy of the photon? What is the wavelength of the photon? Use a value of 4.14 times 10 to the negative 15 electron volt seconds for the value of the Planck constant. Give your answer to the nearest nanometer.

Let’s begin with the first part of the question. Here we have an electron transitioning down an energy level, so energy must be transferred out of it. This energy leaves the electron via the photon. Recall that the energy difference between the electron’s initial and final levels, which we can call Δ𝐸, corresponds to or has the same value as the energy of the emitted photon. We can calculate Δ𝐸 by subtracting the binding energy at level two from the binding energy at level three. Let’s substitute in these values from the diagram, and we have Δ𝐸 equals negative 1.51 electron volts minus negative 3.40 electron volts, which equals 1.89 electron volts. This is the energy transferred out of the electron by means of the photon. And therefore, the photon’s energy is 1.89 electron volts.

Now, moving on to the second part of the question, we need to determine the wavelength of the photon. To do this, let’s recall that we can relate the energy 𝐸 of a photon to its wavelength 𝜆 using the formula 𝐸 equals ℎ𝑐 divided by 𝜆, where ℎ is the Planck constant whose value we’ve been given in the question statement and 𝑐 is the speed of light, 3.0 times 10 to the eight meters per second. Since we wanna find the wavelength, let’s rearrange this formula to make 𝜆 the subject. We can do this by multiplying both sides of the formula by 𝜆 over 𝐸. So 𝐸 cancels out of the left-hand side, and 𝜆 cancels out of the right-hand side. Thus, we have 𝜆 equals ℎ𝑐 over 𝐸.

Now, we can substitute in the values for the Planck constant, the speed of light, and the photon’s energy, which we found in the first part of the question. And notice that we can cancel units of electron volts from the numerator and denominator, as well as units of seconds and per seconds in the numerator. This leaves only units of meters, and that’s a good sign because we’re solving for a wavelength which is a distance measurement. And now, calculating, this comes out to 6.57 times 10 to the negative seven meters. All that’s left to do is express this value to the nearest nanometer.

Recall that one nanometer equals 10 to the negative nine meters. So, to convert, we’ll move the decimal point of our value one, two places to the right. Thus, we know that 𝜆 equals 657 times 10 to the negative nine meters, or 657 nanometers. This is the wavelength of the photon.