# 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?

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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

#### 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 thewavelengthof the emittted photon#R# is theRydberg constant, equal to#1.097 * 10^(7)# #"m"^(-1)# #n_f# is thefinal energy levelof the transition#n_i# is theinitial energy levelof 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 theenergyof the photon#h# isPlanck's constant, equal to#6.626 * 10^(-34)color(white)(.)"J s"# #c# is thespeed of lightin 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 **first energy level** of a hydrogen atom, i.e. its **ground state**, is at

If an incoming photon has an energy that is at least *high enough* to be considered outside the influence of the nucleus, and thus outside the atom