. An excited hydrogen atom with an electron in the n 5 5 state emits light having a frequency of 6.90 3 1014 s21. Determine the principal quantum level for the final state in this electronic transition.

2

Explanation:

Given that the energy of a photon of light is obtained from;

ΔE = hf = 6.6 * 10^-34 Js * 6.903 * 10^14 = 45.6 * 10^-20 J

For moving from n=2 -n-=5 the electron absorbed energy hence the energy is positive.

but;

ΔE = -2.18 * 10^-18 (1/n^2final - 1/n^2initial)

45.6 * 10^-20 J= 2.18 * 10^-18 J (1/n^2final - 1/5^2)

0.21 =  (1/n^2final - 1/5^2)

1/n^2final = 0.21 + 1/5^2

n^2final = 4

nfinal = 2