Bohr theory Problem

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  • #1
docnet
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Homework Statement:
What is the ionization energy for an electron of a 1-electron uranium atom?
Relevant Equations:
##\frac{1}{\lambda}=R_H(\frac{1}{n_f^2}-\frac{1}{n_i^2})##
A neutral uranium atom has 92 electrons and 92 protons. in a violent nuuclear event a uranium nucleus is stripped of all 92 electrons. The resulting bare nucleus captures a single free electron from the surroundings. Given that the ionization energy for hydrogen is ##13.6eV##, derive the approximate value for the maximum energy of the photon that can be given off as the nucleus captures its first electron.

Solution:

The energy of the photon is inversely proportional to ##\lambda##

and the wavelength ##\lambda## is related to the mass of the electron and the energy levels of the atom by the equation

##\frac{1}{\lambda}=R_H(\frac{1}{n_f^2}-\frac{1}{n_i^2})##

where ##R_H=\frac{2\pi ^2 me^4}{h^3 c}##

Our mass of the uranium is be approximately 92 times larger than that of the hydrogen atom. This means the ionization energy for the hydrogen, or the energy of the photon given off by the captured electron, is given by

##92\times13.6eV=1251.2eV##
 

Answers and Replies

  • #3
docnet
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Venturing well outside my knowledge here, but at https://en.wikipedia.org/wiki/Energ...y_level:_atom/ion_with_nucleus_+_one_electron I see a formula with Z2, not Z. Is that a worry?

Yes you are right. ##Z^2## definitely changes our solution.

using the formula from wiki for a hydrogen-like atom with atomic number Z,

##\frac{1}{\lambda} =\frac{m_{electron}e^4}{8ε_0^2 h^3 c}Z^2(\frac{1}{n^2_f}-\frac{1}{n^2_f})##

##E= \frac{hc}{\lambda} = \frac{m_{electron}e^4}{8ε_0^2 h^2}(92)^2(1-\frac{1}{∞})\approx 115,000eV##

which is confirmed here
https://www.omnicalculator.com/physics/hydrogen-like-atom (thank you).
 
  • #4
Frodo
Gold Member
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Our mass of the uranium is be approximately 92 times larger than that of the hydrogen atom.
That statement is wrong for mass as U has lots of neutrons as well as protons. Compare the U and H atomic weights of ~238 and ~1.

It is correct for charge when you remove approximately as the U nucleus has 92 protons whereas the H nucleus has only 1 proton.

Mass does not come into your calculation as gravity is so weak in comparison with electric forces. Your calculation depends on charge.
 
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