Question

An electron in the n=6 level of the hydrogen atom relaxes to a lower energy level, emitting light of λ=93.8nm. Find the principal level to which the electron relaxed.

Answer 1

We know that 1/λ = R [ 1/n1^2 - 1/n2^2] , λ = wavelength

R = 1.09678 x 10^-2 nm^-1

Given that electron in the n=6 level of the hydrogen atom relaxes to a lower energy level, emitting light of λ=93.8nm.

i.e. n1= ? , n2= 6

λ = 93.8 nm

Then,

1/λ = R [1/n1^2 - 1/n2^2]

1/ 93.8 nm = (1.09678 x 10^-2 nm^-1) [ 1/n1^2 - 1/6^2]

1/ 93.8 nm  = (1.09678 x 10^-2 nm^-1) [ 1/n1^2 - 1/36]

[ 1/n1^2 - 1/36] = 0.97

[ 1/n1^2 - 1/36] = 0.97

1/n1^2 = 0.02 + 0.97

1/n1^2 = 0.99 = 1

n1^2 = 1

Then, n1 =1

Therefore, to n=1 principal level to which the electron relaxed.