An electron changes from an #n = 2# to an #n = 6# energy state. What is the energy of the photon in joules?
1 Answer
Explanation:
The question wants you to determine the energy that the incoming photon must have in order to allow the electron that absorbs it to jump from
A good starting point here will be to calculate the energy of the photon emitted when the electron falls from
#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)#
Plug in your values to find
#1/lamda = 1.097 * 10^7color(white)(.)"m"^(-1) * (1/2^2 - 1/6^2)#
#1/lamda = 2.4378 * 10^6 color(white)(.)"m"^(-1)#
This means that you have
#lamda = 4.10 * 10^(-7)color(white)(.)"m"#
So, you know that when an electron falls from
To find the energy of this photon, you can use the Planck - Einstein relation, which looks like this
#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 * 10^8 color(white)(.)"m s"^(-1)#
As you can see, this equation shows you that the energy of the photon is inversely proportional to its wavelength, which, of course, implies that it is directly proportional to its frequency.
Plug in the wavelength of the photon in meters to find its energy
#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)))))/(4.10 * 10^(-7) color(red)(cancel(color(black)("m"))))#
#color(darkgreen)(ul(color(black)(E = 4.85 * 10^(-19)color(white)(.)"J")))#
I'll leave the answer rounded to three sig figs.
So, you can say that in a hydrogen atom, an electron located on