6.4 Bohr’s model of the hydrogen atom  (Page 7/37)

Size and ionization energy of the hydrogen atom in an excited state

If a hydrogen atom in the ground state absorbs a 93.7-nm photon, corresponding to a transition line in the Lyman series, how does this affect the atom’s energy and size? How much energy is needed to ionize the atom when it is in this excited state? Give your answers in absolute units, and relative to the ground state.

Strategy

Before the absorption, the atom is in its ground state. This means that the electron transition takes place from the orbit $m=1$ to some higher n th orbit. First, we must determine $n$ for the absorbed wavelength $\lambda =93.7\text{nm}.$ Then, we can use [link] to find the energy $E_n$ of the excited state and its ionization energy $E_{\infty ,n},$ and use [link] to find the radius $r_n$ of the atom in the excited state. To estimate n , we use [link] .

Solution

Substitute $m=1$ and $\lambda =93.7\text{nm}$ in [link] and solve for $n.$ You should not expect to obtain a perfect integer answer because of rounding errors, but your answer will be close to an integer, and you can estimate n by taking the integral part of your answer:

The radius of the $n=6$ orbit is

Thus, after absorbing the 93.7-nm photon, the size of the hydrogen atom in the excited $n=6$ state is 36 times larger than before the absorption, when the atom was in the ground state. The energy of the fifth excited state ( $n=6$ ) is:

After absorbing the 93.7-nm photon, the energy of the hydrogen atom is larger than it was before the absorption. Ionization of the atom when it is in the fifth excited state ( $n=6$ ) requites 36 times less energy than is needed when the atom is in the ground state:

Significance

We can analyze any spectral line in the spectrum of hydrogen in the same way. Thus, the experimental measurements of spectral lines provide us with information about the atomic structure of the hydrogen atom.

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