Question #03297

1 Answer
Nov 21, 2017

Answer:

#n_f = 3#

Explanation:

Your tool of choice here will be the Rydberg equation, which allows you to calculate the wavelength of the photon emitted when an electron in a hydrogen atom makes an #n_i -> n_f# transition.

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

Here

  • #R# is the Rydberg constant, equal to #1.097 * 10^(7)# #"m"^(-1)#
  • #n_i# is the energy level from which the electron falls
  • #n_f# is the energy level to which the electron falls

In your case, you know that the electron falls from the seventh energy level, so

#n_i = 7#

Your goal here is to find the value of the final energy level, #n_f#. Rearrange the Rydberg equation to solve for #n_f#.

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

#n_i^2 * n_f^2 = lamda * R * (n_i^2 - n_f^2)#

#n_i^2 * n_f^2 + lamda * R * n_f^2 = lamda * R * n_i^2#

This is equivalent to

#n_f^2 * (n_i^2 + lamda * R) = lamda * R * n_i^2#

which gets you

#n_f = sqrt( (n_i^2 * lamda * R)/(n_i^2 + lamda * R))#

Now all you have to do is to plug in the values that you have--do not forget that the wavelength of the photon must be expressed in meters!

#n_f = sqrt( (7^2 * 1005 * 10^(-9)color(red)(cancel(color(black)("m"))) * 1.097 * 10^7 color(red)(cancel(color(black)("m"^(-1)))))/(7^2 + 1005 * 10^(-9)color(red)(cancel(color(black)("m"))) * 10.97 * 10^7 color(red)(cancel(color(black)("m"^(-1))))))#

#n_f = 2.99983 ~~ 3#

Therefore, you can say that a photon of wavelength #"1005 nM"# is emitted when an electron in a hydrogen atom falls from #n_i = 7# to #n_f = 3#, a transition that is part of the Paschen series.

https://thecuriousastronomer.wordpress.com/2013/08/20/emission-line-spectra/