Question

An electron in an excited state of a hydrogen atom emits two photons in succession, the first at 2624 nm and the second at 97.20 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

The equations you need are:

E = hc/λ and

Here,

E = Energy

h = plank’s constant = 6.626 x 10³⁴ Js

c = velocity of light = 3 x 10⁸ m/s

λ = wavelength

For 2624 nm = 2624 x 10^-9 m

ΔE1 = (6.626 x 10^-34 Js x 3 x 10⁸ m/s) / (2624 x 10^-9 m)

       = 7.58 x 10^-20 J

Now,

ΔE = 2.18 x 10^-18J ( 1/n_f² - 1/n_i² )

7.58 x 10^-20 J = 2.18 x 10^-18 J (1/4² - 1/ n_i²)

0.035 = 1/16 - 1/ n_i²

0.035 = 0.0625 - 1/ n_i²

1/ n_i² = 0.0275

n_i² = 36

n_i = 6

For 97.20 nm = 97.20 x 10^-9m

ΔE1 = (6.626 x 10^-34 Js x 3 x 10⁸ m/s) / (97.20 x 10^-9 m)

       = 2.05 x 10^-18 J

Now,

ΔE = 2.18 x 10^-18J ( 1/n_f² - 1/n_i² )

2.05 x 10^-18 J = 2.18 x 10^-18 J (1/1 - 1/ n_i²)

0.940 = 1 - 1/ n_i²)

1/ n_i² = 0.060

n_i² = 16

n_i = 4