1. An electron in an excited state of hydrogen undergoes a transition to the n =
ID: 998243 • Letter: 1
Question
1. An electron in an excited state of hydrogen undergoes a transition to the n = 1 level. In the process, a photon of energy 1.94×10-18 J is emitted.
What was the value of the principal quantum number for the higher energy level?
2. 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?
Explanation / Answer
1. energy dE = 2.18 x 10^-18 [1/nf^2 - 1/ni^2]
nf = 1
we get,
1.94 x 10^-18 = 2.18 x 10^-18[1 - 1/ni^2]
ni = 3
So, the value for principle quantum number for the higher energy level is 3.
2. dE = hc/l = 2.18 x 10^-18[1/nf^2 - 1/ni^2]
l = 97.20 nm = 97.20 x 10^-9 m
nf = 1
6.626 x 10^-34 x 3 x 10^8/97.20 x 10^-9 = 2.18 x 10^-18[1 - 1/ni^2]
ni = 4
Now with nf = 4,
l = 2624 nm = 2624 x 10^-9 m
nf = 1
6.626 x 10^-34 x 3 x 10^8/2624 x 10^-9 = 2.18 x 10^-18[1/16 - 1/ni^2]
ni = 6
So the principle quantum number of the initial excited state is 6, and that os intermediate state is 4.
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