### Video Transcript

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 couldโve occupied before being ionized?

We can record the energy of an electron in the ground state of a hydrogen atom as ๐ธ sub zero and we can write the wavelength of the photon thatโs absorbed by the atom as ๐. We want to solve for the lowest possible energy level of the electron that is ionized by absorbing this photon. Weโll call that energy level ๐ sub ๐. A sketch of the energy levels of the hydrogen atom can help clarify this situation.

If ๐ธ 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. Thatโs the point at which, if an electronโs energy is greater than or equal to it, itโs free to escape the atom. For hydrogen, weโre told that the ground state energy of an electron is 13.6 electron volts below the ionization energy. Thatโs the reason for the negative sign. Weโre told in the problem statement about a photon thatโs incident on an electron in some energy level, where the photon has a wavelength ๐ and is energetic enough to ionize the electron. That is, make it completely escape that particular hydrogen atom. Given ๐, we want to solve for the lowest possible energy level that the ionized electron couldโve occupied.

To figure out that lowest possible energy level, letโs recall that photon energy is equal to Planckโs constant times frequency or Planckโs constant, โ, times ๐ over ๐. We can also recall that for a hydrogen atom, the energy of the ๐th energy level equals the energy of the ground state divided by ๐ squared. Going back to our photon energy equation, if we call ๐ธ sub ๐, the energy of the incident photon, we can represent that energy graphically on our diagram like this. The horizontal line labelled ๐ธ sub ๐ 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, ๐ธ sub ๐, would be given enough energy by the incident photon in order to ionize an electron.

So all the hydrogen atom energy levels above ๐ธ sub ๐ are candidates for our solution ๐ sub ๐. Based on our diagram, itโs the energy of the hydrogen atom thatโs directly above ๐ธ sub ๐, which is that lowest possible ionizable ๐-value we want to solve for. To figure that integer out, weโll take our photon energy and set it equal to the magnitude of our hydrogen atom energy-level energy.

Now weโve said these two energy levels exactly equal to one another, but we recognize that probably they are not exactly equal. But in fact, thereโs a gap between ๐ธ sub ๐ and ๐ sub ๐. So when we solve for ๐ in this equation, we donโt expect an integer but whatever value we do get, if itโs not an integer, we can round that up to the next highest integer. And that will be ๐ sub ๐. So letโs solve for ๐ using this expression.

When we rearrange it, we find that ๐ equals the square root of the magnitude of ๐ธ zero times ๐ over โ๐. โ, Planckโs constant, we take to be exactly 6.626 times 10 to the negative 34th joule seconds. And ๐, the speed of light, we take to be 3.00 times 10 to the eighth meters per second. Weโve been given ๐ธ sub zero in the problem statement as well as ๐. And now that we know โ and ๐, weโre ready to plug in for the values in this equation.

When we do plug these numbers in, notice two things about how weโve written them. First, weโve expressed the wavelength ๐ in units of meters. And second, weโve 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 ๐ using this expression which remember, may not be an integer, we find itโs equal to 2.77. So indeed, there really is a gap between ๐ธ sub ๐ 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 ๐, to get out of it ๐ sub ๐, the integer we do want, we simply round up to the next highest integer. So itโs the ๐ equals three energy level of the hydrogen atom, which is the lowest possible energy level the electrons of which this photon can ionize.