Question

Determine which H atom transition (ninitial,nfinal) was responsible for the emission of a photon with a frequency of 6.91 x 1014 Hz.
[RH = 2.18 x 10-18 J; h = 6.63 x 10-34 J s ]

Answer 1

The answer is: (ninitial,nfinal) = (5,2)

The transition is from n(initial) = 5 to n(final) = 2

See solution below:

Wavelength λ = c/f

= 2.998 x 10⁸/6.91 x 10¹⁴

= 4.34 x 10^-7 m = 434 nm

where c is speed of light and f is frequency

Since λ falls in the visible region, the transition comes from the Balmer series and n(final) = 2

Rydberg equation:

E = hf = RH(1/(n(final)² - 1/n(initial)²)

where RH is Rydberg constant and h is Planck constant

6.63 x 10^-34 x 6.91 x 10¹⁴ = 2.18 x 10^-18 x (1/2² - 1/n(initial)²)

1/n(initial)² = 0.039974

n(initial)² = 1/0.039974 = 25

n(initial) = 5

Thus the transition is from n(initial) = 5 to n(final) = 2

i.e. (ninitial,nfinal) = (5,2)