If light of wavelength #"434 nm"# irradiates a hydrogen atom, what energy level transition will it make? What series is this in the emission line spectrum?

1 Answer
Jun 11, 2017

#n=5 -> n=2#

Explanation:

All you have to do here is to use the Rydberg formula for a hydrogen atom

#1/(lamda_"e") = R * (1/n_1^2 - 1/n_2^2)#

Here

  • #lamda_"e"# is the wavelength of the emitted photon (in a vacuum)
  • #R# is the Rydberg constant, equal to #1.097 * 10^(7)# #"m"^(-1)#
  • #n_1# represents the principal quantum number of the orbital that is lower in energy
  • #n_2# represents the principal quantum number of the orbital that is higher in energy

Notice that you need to have #n_1 < n_2# in order to avoid getting a negative value for the wavelength of the emitted photon.

So, convert the wavelength from nanometers to meters

#434 color(red)(cancel(color(black)("nm"))) * "1 m"/(10^9color(red)(cancel(color(black)("nm")))) = 4.34 * 10^(-7)# #"m"#

Rearrange the Rydberg formula to isolate the unknown variables

#1/n_1^2 - 1/n_2^2 = 1/lamda_"e" * 1/R#

Plug in your values to find

#1/n_1^2 - 1/n_2^2 = 1/(4.34 * color(blue)(cancel(color(black)(10^(-7))))color(red)(cancel(color(black)("m"))) * 1.097 * color(blue)(cancel(color(black)(10^7)))color(red)(cancel(color(black)("m"^(-1)))))#

#1/n_1^2 - 1/n_2^2 = 0.21#

This will be equivalent to

#(n_2^2 - n_1^2)/(n_1 * n_2)^2 = 21/100#

At this point, you have two equations with two unknowns, since

#{(n_2^2 - n_1^2 = 21), (n_1 * n_2 = sqrt(100)) :}#

Use the second equation to write

#n_1 = 10/n_2" "color(darkorange)("(*)")#

Plug this into the first equation to get

#n_2^2 - (10/n_2)^2 = 21#

#n_2^2 - 100/n_2^2 = 21#

This is equivalent to

#n_2^4 - 100 = 21 * n_2^2#

#n_2^4 - 21n_2^2 - 100 = 0#

If you take #n_2^2 = x#, you will have

#x^2 - 21x - 100 = 0#

Use the quadratic equation to get

#x_(1,2) = ( -(-21) +- sqrt( (-21)^2 - 4 * 1 * (-100)))/(2 * 1)#

#x_(1,2) = (21 +- sqrt(841))/2#

#x_(1,2) = (21 +- 29)/2 implies {( x_1 = (21 - 29)/2 = -4), (x_2 = (21 + 29)/2 = 25) :}#

Since we know that

#n_2^2 = x#

you can only use the positive solution here, so

#n_2^2 = 25 implies n_2 = 5#

According to equation #color(darkorange)("(*)")#, you will have

#n_1 = 10/5 = 2#

This means that your transition is taking place from

#n=5 -> n=2#

which is part of the Balmer series.

http://www.daviddarling.info/encyclopedia/B/Balmer_series.html