Question

A hydrogen atom in the ground state absorbs a 12.75 eV photon. Immediately after the absorption, the atom undergoes a quantum jump to the next-lowest energy level.

What is the wavelength of the photon emitted in this quantum jump?

Express your answer using four significant figures.

I've seen this question before, but I'm looking the wavelength, not
the energy, or n. Thanks!

Answer #1

**For the hydrogen atom the energy of the quantum state n
is E = - (13.6 eV) / n^2. The ground state has n = 1 and E = - 13.6
eV.
The energy of the atom after absorption is -13.6 + 12.75 = -0.85
eV. The quantum number for this energy state is given by
-0.85 = -13.6/n^2
n^2 = 13.6/0.85 = 16
n =4.
The next-lower energy level has energy
E = -13.6/3^2 = -1.511
so the transition emits a photon of energy -0.85-(-1.511) = 0.661
eV**

**0.661 = hc / lambda**

**lambda = 6.63x10^-34 Js x 3.00x10^8 m/s? / 0.661 *
10^-19 = 3.02*10^-6 m**

A hydrogen atom is initially at the ground state and then
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(a) How many possible wavelengths will be emitted as the atom
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Answer: (number) ___________
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A hydrogen atom is in its ground state
(ni = 1)
when a photon impinges upon it.
The atom absorbs the photon, which has precisely the energy
required to raise the atom to the
nf = 3
state.
(a)
What was the photon's energy (in eV)?
eV
(b)
Later, the atom returns to the ground state, emitting one or
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to illustrate your answer)
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A
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I got Nf=1.14 and it wasn't right

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