How do you calculate the wavelength of the light emitted by a hydrogen atom during a transition of its electron from the n = 4 to the n = 1 principal energy level? Recall that for hydrogen #E_n = -2.18 xx 10^-18 J(1/n^2)#

1 Answer
Truong-Son N.
Dec 16, 2016

Note that you were given only one energy state. If you consider two energy states, from #n = 4# to #n = 1#, we have:

#E_1 - E_4 = color(blue)(DeltaE)#

#= -2.18xx10^(-18) "J"(1/n_f^2 - 1/n_i^2)#

#= -2.18xx10^(-18) "J"(1/1^2 - 1/4^2)#

#= -2.18xx10^(-18) "J"(15/16)#

#=# #-color(blue)(2.04xx10^(-18) "J")#

After you obtain the energy, then you can realize that that energy has to correspond exactly to the energy of the photon that came in:

#|DeltaE| = E_"photon" = hnu = (hc)/lambda#

where #h# is Planck's constant, #c# is the speed of light, and #lambda# is the wavelength of the incoming photon. Thus, the wavelength is:

#=> color(blue)(lambda) = (hc)/(E_"photon") = ((6.626xx10^(-34) "J"cdot"s")(2.998xx10^(8) "m/s"))/(2.04xx10^(-18) "J")#

#= 9.720 xx 10^(-8)# #"m"#

#=# #color(blue)("97.20 nm")#

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