Video Transcript
The diagram shows the transition of an electron in a hydrogen atom from 𝑛 equals three to 𝑛 equals one, emitting a photon as it does so. What is the energy of the photon? Give your answer to two decimal places. What is the frequency of the photon? Use a value of 4.14 times 10 to the negative 15 electron volt seconds for the Planck constant. Give your answer in scientific notation to two decimal places.
Here we have an electron in a downward transition. So energy is transferred away from the electron by means of a photon. Let’s begin with the first part of this question and determine the photon’s energy. To do this, we should recall that the difference between the electron’s initial and final energy levels, which we call Δ𝐸, corresponds to or has the same value as the emitted photon.
The diagram tells us the binding energies at the electron’s initial and final levels. So substituting them into the formula for Δ𝐸, we have negative 1.51 electron volts minus negative 13.6 electron volts, which gives us 12.09 electron volts. This value is already written to two decimal places, so we have our answer. The energy of the emitted photon is 12.09 electron volts.
Moving on to the next part of the question, we need to find the photon’s frequency. Let’s recall that we can relate to photon’s energy 𝐸 to its frequency 𝑓 using the formula 𝐸 equals ℎ times 𝑓, where ℎ is the Planck constant whose value has been given to us. Because we want to find the frequency of the photon, let’s copy the formula and divide both sides by ℎ so that we can cancel that term from the right-hand side, leaving 𝑓 by itself as the subject of the formula.
Now flipping it the other way and writing it a bit more neatly, we have that the photon’s frequency equals its energy divided by the Planck constant. And since we already know both of those values, let’s substitute them into the formula. We have 12.09 electron volts divided by 4.14 times 10 to the negative 15 electron volt seconds. And notice that we can cancel units of electron volts from the numerator and denominator, leaving only units of seconds in the denominator or units of per seconds for the final answer. This is a good sign because per seconds or hertz is the SI unit of frequency.
Now that we’ve figured out the units, let’s calculate the final value by dividing 12.09 by 4.14 times 10 to the negative 15, which comes out to about 2.9203 times 10 to the 15 hertz. And finally, rounding our answer to two decimal places, we found that the frequency of the emitted photon is 2.92 times 10 to the 15 hertz.
