Question

For the Lyman series, `n_f = 1`. How can we calculate the `E_"photon"` for the bandhead of the Lyman series (the transition `n = oo -> n = 1` for emission) in joules and in eV?

The same energy is needed for the transition `n = 1 -> n = oo`, which is the ionization potential for a hydrogen atom.

Answer 1

`2.17 * 10^(-18)` `"J"`

`"13.6 eV"`

Explanation:

Your tool of choice here will be the Rydberg equation

`1/(lamda) = R * (1/n_f^2 - 1/n_i^2)`

Here

• `lamda` si the wavelength of the emittted photon
• `R` is the Rydberg constant, equal to `1.097 * 10^(7)` `"m"^(-1)`
• `n_f` is the final energy level of the transition
• `n_i` is the initial energy level of the transition

In your case, you have the

`n_i = oo -> n_f = 1`

transition, which is part of the Lyman series. The first thing to notice here is that when `n_i = oo`

`1/n_i^2 -> 0`

which implies that the Rydberg equation can be simplified to this form

`1/lamda = R * (1/1^2 - 0)`

`1/(lamda) = R`

You can thus say that the wavelength of the emitted photon will be equal to

`lamda = 1/R`

`lamda = 1/(1.097 * 10^7color(white)(.)"m"^(-1)`

`lamda = 9.158 * 10^(-8)color(white)(.)"m"`

Now, to find the energy of the photon emitted during this transition, you can use the Planck - Einstein relation

`E = (h * c)/lamda`

Here

• `E` is the energy of the photon
• `h` is Planck's constant, equal to `6.626 * 10^(-34)color(white)(.)"J s"`
• `c` is the speed of light in a vacuum, usually given as `3 * 10^8color(white)(.)"m s"^(-1)`

Plug in your value to find

`E = (6.626 * 10^(-34)color(white)(.)"J" color(red)(cancel(color(black)("s"))) * 3 * 10^8 color(red)(cancel(color(black)("m"))) color(red)(cancel(color(black)("s"^(-1)))))/(9.158 * 10^(-8)color(red)(cancel(color(black)("m"))))`

`color(darkgreen)(ul(color(black)(E = 2.17 * 10^(-18)color(white)(.)"J")))`

I'll leave the answer rounded to three sig figs.

To convert this to electronvolts, use the fact that

`"1 eV" = 1.6 * 10^(-19)color(white)(.)"J"`

You will end up with

`2.17 * 10^(-18) color(red)(cancel(color(black)("J"))) * "1 eV"/(1.6 * 10^(-19)color(red)(cancel(color(black)("J")))) = color(darkgreen)(ul(color(black)("13.6 eV")))`

This basically means that you need `"13.6 eV"` to ionize a hydrogen atom. In other words, the first energy level of a hydrogen atom, i.e. its ground state, is at `-"13.6 eV"`.

https://chemistry.stackexchange.com/questions/44625/what-is-the-series-limit-in-a-spectral-line

If an incoming photon has an energy that is at least `"13.6 eV"`, then the electron will move to an energy level that is high enough to be considered outside the influence of the nucleus, and thus outside the atom `->` the hydrogen atom is ionized to a hydrogen cation, `"H"^(+)`.