Question

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)`

Answer 1

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")`