Quantum Electron Energy Level for Hydrogenic atom

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The discussion centers on calculating the maximum wavelength of light required to ionize a doubly ionized Lithium atom (Li++), which has one electron around a +3 charge nucleus. The energy of the atom is derived from the formula for hydrogen, adjusted for the increased nuclear charge, leading to an expected energy of -40.818 eV for the ground state. An initial calculation yielded a wavelength of 30.38 nm, but the correct wavelength is actually 10.1 nm, indicating a misunderstanding of mass contributions in the energy formula. It was clarified that 'm' refers to the mass of the electron, and the charge term involves the nucleus's charge squared, explaining the discrepancy. The discussion concludes with a resolution of the misunderstanding regarding the factors affecting the energy levels of hydrogenic atoms.
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1. Consider doubly ionized Lithium (Li++), which has one electron orbiting a +3 charge nucleus. Assuming the electron is in the ground state (n=1), what is the maximum wavelength of light, λ, that would completely ionize the Li++? (free the electron from the nucleus),
all variables and given/known data

2. Energy of a Hydrogen atom is \frac{-13.606eV}{n^{2}} where 'n' is the energy level. The general formula is:
E = \frac{-1}{4\pi r^{2}}\frac{me^{4}}{2\bar{h}n^{2}}
for atoms with one electron.

For light:
E = \frac{1240eV-nm}{λ}

3. I assumed that Li++ would be like a hydrogen atom with 3 times the mass. Since E is proportional to m, the energy of Li should be -13.606eV*(3)/n^2 = 40.818eV. (n=1) I plugged this into the equation λ = 1240eV-nm/E, and got λ = 30.38nm.
The correct answer is 10.1nm... so I am off by a factor of three. So does this mean that the doubly ionized lithium actually has 9 times the mass?

Thank you!
 
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Mass is not the only difference between H and Li.
 
In the energy formula, does m stand for the mass of the atom or the mass of the electron?
 
Hi guys,
I figured it out. As it turns out m is mass of the electron, and e^4 is actually the charge of the nucleus squared times the charge of the electron squared. Which happens to be e^4 in an H atom. In this case it was e^2*(3e)^2 That's where the factor of 9 comes from! Thanks
 
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