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.
The same energy is needed for the transition
1 Answer
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
#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
If an incoming photon has an energy that is at least