Question

a ground state hydrogen atom absorbs a photon of light having a wavelength of 93.73 nm. it then gives off a photon having a wavelength of 410.1 nm. what is the final state of the hydrogen atom?

Answer 1

Energy absorbed = h*c [1/lambda1 - 1/lambda2]

                                =(6.626*10^-34 * 3*10^8) * [1/(93.73*10^-9) - 1/(410.1*10^-9)]

                                = 1.9878*10^-25 * (8.23*10^6)

                                = 1.64*10^-18 J

                                 = 1.64*10^-18/ (1.6*10^-19) eV

                                = 10.22 eV

Now for H atom:

E absorbed =13.6 [1/ni^2 - 1/nf^2]

10.22 = 13.6* [1- 1/nf^2]

[1- 1/nf^2] =0.752

1/nf^2 = 0.25

nf^2 = 4

nf = 2

So, final state is 2