Exploring the Bohr Model: Visible Light Transitions in Li++

[ChiefKeeper92]

Bohr Model
1. Assuming that the wavelengths of visible light lie between about 300nm and 700nm, what transitions in Li++ (hydrogenic lithium ions, Z=3) would be visible. Identify each transition by initial and final principal quantum number n. Also identify those transitions that are also seen in hydrogen (3->2, 4->2, 5->2 in hydrogen)

2. E=13.6eV*(Z^2)*((1/n)^2-(1/m)^2)
λ=1240/E


3. I ran through it but I keep getting the same wavelengths as hydrogen. I don't know what I'm doing wrong. Do I multiply n*Z on the bottom before I square it?

 

[ehild]

What did you use for Z? Remember, the atomic number of Lithium is 3, and the nucleus of the Li++ ion has 3e charge.

ehild

 

[ChiefKeeper92]

I used Z=3 but do I need to multiply Z by n and m?

 

[ehild]

I do not understand you. You need to substitute different n-s and m-s into the formula. First evaluate the expression in the parentheses, (1/n²-1/m²). Then multiply the result by (13.6 * 9).

ehild

 

[asteriatic]

I would like to address your questions and concerns regarding the Bohr Model and the visible light transitions in Li++. First, let's review the Bohr Model and its application to hydrogen atoms. The Bohr Model is a simplified representation of the atom where electrons orbit around the nucleus in specific energy levels or shells. These energy levels are represented by the principal quantum number, n, and the energy of each level is given by the formula E = -13.6 eV/n^2, where eV is the unit for energy and n is the principal quantum number.

Now, in the case of Li++, we have a hydrogenic ion with a nuclear charge of Z=3. This means that the energy levels will be slightly different compared to hydrogen. Using the formula provided in your question, we can calculate the energy of each transition in Li++ and determine which ones fall within the visible light range (300nm-700nm).

For the first transition, we have n=3 to n=2, which corresponds to an energy of 16.2 eV. Using the formula, we can calculate the wavelength of this transition to be approximately 76.5 nm. This falls outside of the visible light range and is therefore not visible.

The second transition, n=4 to n=2, has an energy of 7.2 eV and a wavelength of 172.2 nm, also outside of the visible light range. However, this transition is also seen in hydrogen and is known as the Lyman series.

Lastly, the third transition, n=5 to n=2, has an energy of 4.8 eV and a wavelength of 258.3 nm, still outside of the visible light range. This transition is also seen in hydrogen and is known as the Balmer series.

Therefore, there are no visible light transitions in Li++ that are unique to this ion. All visible light transitions in Li++ are also seen in hydrogen. This is because the Bohr Model is a simplified representation and does not take into account the effects of multiple electrons and their interactions.

In regards to your third question, the formula provided already takes into account the nuclear charge, Z. So you do not need to multiply n*Z on the bottom before squaring it. I hope this clarifies your doubts and helps you with your calculations. Keep exploring and asking questions!