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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.

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##

**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##