Question

An electron in an excited state of a hydrogen atom emits two photons in succession, the first at 3037 nm and the second at 94.92 nm, to return to the ground state (n=1). For a given transition, the wavelength of the emitted photon corresponds to the difference in energy between the two energy levels.

What were the principal quantum numbers of the initial and intermediate excited states involved?

Answer 1

dE1 = hc/l = 2.18 x 10^-18[1/n1^2-1/n2^2]

where,

h = 6.626 x 10^-34 J.s

c = 3 x 10^8 m.s-1

l = 94.92 nm = 9.492 x 10^-8 m

So,

dE2 = 6.626 x 10^-34 x 3 x 10^8/9.492 x 10^-8 = 2.094 x 10^-18 J

Feed values with n1 = 1,

2.094 x 10^-18 = 2.18 x 10^-18[1-1/n2^2]

n2 = 5

Now,

l = 3037 nm = 3.037 x 10^-6 m

Feed values,

dE1 = 6.626 x 10^-34 x 3 x 10^8/3.037 x 10^-6 = 6.55 x 10^-20 J

So, with n2 = 5

dE1 = 2.18 x 10^-18[1/n1^2-1/n2^2]

6.55 x 10^-20 = 2.18 x 10^-18 [1/n1^2-1/25]

n1 = 4

So the principle quantum numbers for the transitions are,

n = 5 to n = 4 to n = 1