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.

1 Answer
Nov 25, 2017

Answer:

#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"^(+)#.