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JewishDude

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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

all variables and given/known data

**2. Energy of a Hydrogen atom is [itex]\frac{-13.606eV}{n^{2}}[/itex] where 'n' is the energy level. The general formula is:**

E = [itex]\frac{-1}{4\pi r^{2}}[/itex][itex]\frac{me^{4}}{2\bar{h}n^{2}}[/itex]

for atoms with one electron.

For light:

E = [itex]\frac{1240eV-nm}{λ}[/itex]

E = [itex]\frac{-1}{4\pi r^{2}}[/itex][itex]\frac{me^{4}}{2\bar{h}n^{2}}[/itex]

for atoms with one electron.

For light:

E = [itex]\frac{1240eV-nm}{λ}[/itex]

**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!

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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