Quantum Electron Energy Level for Hydrogenic atom

[JewishDude]

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!

 

[voko]

Mass is not the only difference between H and Li.

 

[TSny]

In the energy formula, does m stand for the mass of the atom or the mass of the electron?

 

[JewishDude]

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

 

[Nickie]

I would like to clarify a few things about the content and provide a response to the given question.

Firstly, the energy levels of a hydrogen atom and a hydrogenic atom (such as Li++) are different. The energy levels of a hydrogenic atom are given by the equation E = -13.606eV/n^2Z^2, where Z is the atomic number and n is the energy level. In the case of Li++, Z=3, so the energy level would be -13.606eV/1^2(3)^2 = -1.5118eV.

Secondly, the formula E = -1/4πr^2(me^4/2ħ^2)n^2 is for the energy of a single electron in a hydrogenic atom, not for the entire atom. So, it cannot be used to calculate the energy of Li++.

Thirdly, the energy of a photon of light is given by the equation E = hc/λ, where h is Planck's constant, c is the speed of light, and λ is the wavelength of light. The equation E = 1240eV-nm/λ is not applicable in this case.

Now, coming to the given question, the maximum wavelength of light that would completely ionize Li++ (remove the electron from the nucleus) can be calculated using the energy difference between the ground state (n=1) and the ionization energy of Li++. The ionization energy of Li++ is given by the difference in energy between the ground state (n=1) and the first excited state (n=2), which is -1.5118eV - (-0.37545eV) = -1.13635eV.

Using the equation E = hc/λ, we can calculate the maximum wavelength of light as λ = hc/E = (6.626x10^-34 Js)(3x10^8 m/s)/1.13635eV = 5.507x10^-7 m = 550.7 nm.

In conclusion, the maximum wavelength of light that would completely ionize Li++ is 550.7 nm. The discrepancy in the given answer of 10.1 nm is due to incorrect use of equations and not considering the difference in energy levels between a hydrogen atom and a hydrogenic atom.