Question #5dc82

1 Answer
Oct 6, 2017

Answer:

Here's what I got.

Explanation:

The first thing that you need to do here is to calculate the wavelength of the photon emitted when your electron undergoes a #n_i = 5 -> n_f = 2# transition in a hydrogen atom.

The wavelength of the photon will then help you determine its energy.

So, you know that when an electron in a hydrogen atom makes a transition from an initial energy level #n_i# to a final energy level #n_f#, it emits a photon whose wavelength is equal to

#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)#

You can rearrange the above equation, which is called the Rydberg equation, as

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

which gets you

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

In your case, you have #n_i = 5# and #n_f = 2#, which means that the emitted photon will have a wavelength of

#lamda = 1/(1.097 * 10^7color(white)(.)"m"^(-1)) * (5^2 * 2^2)/(5^2 - 2^2)#

#lamda = 4.34 * 10^(-7)color(white)(.)"m"#

Expressed in nanometers--recall that #"1 m" = 10^9# #"nm"#--the wavelength of the emitted photon will be

#lamda = color(darkgreen)(ul(color(black)("434 nm")))#

Now, in order to find the energy of the photon, you need to use a variation of 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)# #"J s"#
  • #c# is the speed of light in a vacuum, usually given as #3 * 10^8# #"m s"^(-1)#

Plug in your value to find

#E = 6.626 * 10^(-34)"J" color(red)(cancel(color(black)("s"))) * (3 * 10^8 color(red)(cancel(color(black)("m"))) color(red)(cancel(color(black)("s"^(-1)))))/(4.34 * 10^(-7) color(red)(cancel(color(black)("m"))))#

#E = color(darkgreen)(ul(color(black)(4.58 * 10^(-19)color(white)(.)"J")))#

I'll leave both answers rounded to three sig figs.

It's worth mentioning that the #n_i = 5 -> n_f = 2# transition, which is part of the Balmer series, is located in the visible portion of the EM spectrum.

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