Orbital energies in single-electron atoms or ions, such as He+, can be described with an equation similar to the Balmer–Rydberg equation:

where Z is the atomic number. What wavelength of light in nanometers is emitted when the electron in He+ falls from n = 3 to n = 2?

Video transcript

Hey everyone, we're told an equation resembling the Balmer Rydberg equation shown below where Z is the atomic number? Can be used to describe the orbital energies of single electron atoms or ions such as lithium. What wavelength of light and nanometers is released? When the electron in lithium falls from N equals four to n equals two. First. Let's recall our Rydberg constant which is going to be 1.97 times to the negative two nanometers to the negative first. And since our lithium falls from N equals forward to N equals two, this means R N final is going to be two and R N initial is going to be four. And when we look at our periodic table, the Z. Which is our atomic number of lithium is going to be three. Now let's go ahead and plug in our values. So we have one over our wavelength equals Z squared, which is three squared times are right for constant and nanometers times one over NF squared, which will be two squared -1 over an initial squared, which is going to be four squared. When we calculate this out, we end up with a one over wavelength of 0. nanometers to the negative first, Converting this into our wavelength, we end up with 54.0194 nm and this is going to be our final answer. Now I hope that made sense. And let us know if you have any questions