Question Video: Using Bohr’s Model to Calculate the Energy of the Photon Produced by the Transition of an Electron from 𝑛 = 2 to 𝑛 = 1 in Li²⁺ Ion

An electron in aLi²⁺ ion moves from the 𝑛 = 2 energy level to the 𝑛 = 1 energy level. Using the Bohr model, calculate, to 4 significant figures, the energy of the photon produced by this transition.

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Video Transcript

An electron in an Li²⁺ ion moves from the 𝑛 equals two energy level to the 𝑛 equals one energy level. Using the Bohr model, calculate, to four significant figures, the energy of the photon produced by this transition.

Li²⁺ is lithium two plus. Lithium has an atomic number of three. So, the lithium nucleus contains three protons. We can take away the charge two plus from the number of protons to work out the number of electrons. In this case, we have a single electron. For this question, we don’t need to worry about the neutrons. In the Bohr model, the only thing that matters is the charges of the nucleus and the electron.

In the Bohr model, the nucleus is surrounded by concentric shells. Electrons are said to occupy these shells at fixed restricted distances from the nucleus. Because of the way this system is sometimes drawn, the Bohr model is sometimes called the planetary model. However, electrons aren’t like planets. They’re more like waves than particles. And the shells are not rings; they are surfaces of spheres.

Each shell is assigned a number, with the number one given to the shell closest to the nucleus. The lower the value of 𝑛, the more stable the electron will be if inside that shell. Each shell has an energy, but the zero point is placed where the electron is infinitely far from the nucleus. This is the equivalent of a shell number of infinity. And it’s the equivalent, in this case, of having a lithium three plus nucleus and a completely separated electron.

At the other end of the spectrum, we have the inner shell, where 𝑛 equals one, where the energy is considered negative. When an electron moves from a high energy level to a low energy level, the difference in energy is released in the form of a photon. In this question, we’re being asked to work out the energy of the photon released when an electron moves from the 𝑛 equals two energy level to the 𝑛 equals one energy level.

So, we’re going from the first excited state of lithium two plus to the ground state, releasing a photon of energy Eph. Before we go any further, we’re going to need to work out the energy of a shell based on its shell number. This is the equation from the Bohr model that tells you the energy of a shell based on the shell number. 𝑍 is equal to the number of protons in the nucleus, while 𝑛 is equal to the shell number.

But what about this term here? Well, the full expression for the energy of any given shell is quite a complex combination of many constants of nature. In this expression, we’re combining the Coulomb constant, the elementary charge, the electron mass, and the reduced Planck constant. But this is a constant times, a constant divided by another constant. It all resolves to a constant. And that constant has a value of about 13.6 electron volts.

If you take the values of the constants in their SI units, you’ll get an answer in joules. And you can convert to electron volts by dividing the value in joules by the elementary charge. Thankfully, all that work has been done, so we have our single condensed constant term of negative 13.6057 electron volts. Now, all we need to do is work out the values for 𝐸 two and 𝐸 one and work out the difference and calculate the energy of the released Photon.

The energy of our core shell is equal to minus 13.6057 electron volts multiplied by three squared over one squared. 𝑍 is the number of protons in the lithium nucleus, which is three. And 𝑛 equals one. This evaluates to minus 122.451 electron volts. The value of our second energy level, 𝐸 two, is equal to negative 13.6057 electron volts multiplied by three squared divided by two squared, which gives us negative 30.6128 electron volts.

Now, we can put those values into our diagram and work out the difference in energy. Our difference in energy is the final energy minus the initial energy, which is equal to negative 122.451 electron volts minus negative 30.6128 electron volts, giving us a change in energy of minus 91.8382 electron volts. So, this is the amount of energy that the electron has lost, becoming more stable. So, the energy of the photon is the negative of this energy, the exact opposite. Because energy can neither be created nor destroyed. Giving us an energy for the photon of positive 91.8382 electron volts.

We can also express this value in joules, which is 1.47141 times 10 to the minus 17 joules. We convert from electron volts to joules by multiplying by 1.60218 times 10 to the minus 19 joules per electron volt. The question asks for our answer to be to four significant figures, which is 91.84 electron volts, or 1.471 times 10 to the minus 17 joules. So, we’ve worked out that when an electron moves from the 𝑛 equals two energy level to the 𝑛 equals one energy level, in the Bohr model, the energy of the photon produced is 1.471 times 10 to the minus 17 joules.

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