The energy of an electron in its ground state in a hydrogen atom is negative 13.6 electron volts. A photon of wavelength 700 nanometers is incident on a hydrogen atom. When this photon is absorbed, the atom becomes ionized. What is the lowest energy level that the electron could’ve occupied before being ionized?
We can record the energy of an electron in the ground state of a hydrogen atom as 𝐸 sub zero and we can write the wavelength of the photon that’s absorbed by the atom as 𝜆. We want to solve for the lowest possible energy level of the electron that is ionized by absorbing this photon. We’ll call that energy level 𝑛 sub 𝑚. A sketch of the energy levels of the hydrogen atom can help clarify this situation.
If 𝐸 sub zero is the energy of the ground state of the hydrogen atom, we know that if we work up to higher energy levels, eventually we get to the ionization level. That’s the point at which, if an electron’s energy is greater than or equal to it, it’s free to escape the atom. For hydrogen, we’re told that the ground state energy of an electron is 13.6 electron volts below the ionization energy. That’s the reason for the negative sign. We’re told in the problem statement about a photon that’s incident on an electron in some energy level, where the photon has a wavelength 𝜆 and is energetic enough to ionize the electron. That is, make it completely escape that particular hydrogen atom. Given 𝜆, we want to solve for the lowest possible energy level that the ionized electron could’ve occupied.
To figure out that lowest possible energy level, let’s recall that photon energy is equal to Planck’s constant times frequency or Planck’s constant, ℎ, times 𝑐 over 𝜆. We can also recall that for a hydrogen atom, the energy of the 𝑛th energy level equals the energy of the ground state divided by 𝑛 squared. Going back to our photon energy equation, if we call 𝐸 sub 𝑝, the energy of the incident photon, we can represent that energy graphically on our diagram like this. The horizontal line labelled 𝐸 sub 𝑝 represents the distance between the ionization energy for a hydrogen atom and the energy of the incident photon. Any hydrogen atom energy level within this range, 𝐸 sub 𝑝, would be given enough energy by the incident photon in order to ionize an electron.
So all the hydrogen atom energy levels above 𝐸 sub 𝑝 are candidates for our solution 𝑛 sub 𝑚. Based on our diagram, it’s the energy of the hydrogen atom that’s directly above 𝐸 sub 𝑝, which is that lowest possible ionizable 𝑛-value we want to solve for. To figure that integer out, we’ll take our photon energy and set it equal to the magnitude of our hydrogen atom energy-level energy.
Now we’ve said these two energy levels exactly equal to one another, but we recognize that probably they are not exactly equal. But in fact, there’s a gap between 𝐸 sub 𝑝 and 𝑛 sub 𝑚. So when we solve for 𝑛 in this equation, we don’t expect an integer but whatever value we do get, if it’s not an integer, we can round that up to the next highest integer. And that will be 𝑛 sub 𝑚. So let’s solve for 𝑛 using this expression.
When we rearrange it, we find that 𝑛 equals the square root of the magnitude of 𝐸 zero times 𝜆 over ℎ𝑐. ℎ, Planck’s constant, we take to be exactly 6.626 times 10 to the negative 34th joule seconds. And 𝑐, the speed of light, we take to be 3.00 times 10 to the eighth meters per second. We’ve been given 𝐸 sub zero in the problem statement as well as 𝜆. And now that we know ℎ and 𝑐, we’re ready to plug in for the values in this equation.
When we do plug these numbers in, notice two things about how we’ve written them. First, we’ve expressed the wavelength 𝜆 in units of meters. And second, we’ve added a energy-conversion factor to change our energy of electron volts and convert it to units of joules. We do this, so the units are consistent all throughout our expression.
When we calculate 𝑛 using this expression which remember, may not be an integer, we find it’s equal to 2.77. So indeed, there really is a gap between 𝐸 sub 𝑝 and the energy level of the lowest possible hydrogen orbital that this photon can excite to ionize the electron in that orbital. What this means is if we take our solved value for 𝑛, to get out of it 𝑛 sub 𝑚, the integer we do want, we simply round up to the next highest integer. So it’s the 𝑛 equals three energy level of the hydrogen atom, which is the lowest possible energy level the electrons of which this photon can ionize.