What is the wavelength of the transition from n=4 to n=3 for Li2+? In what region of the spectrum does this emission occur? Li2+ is a hydrogen-like ion. Such an ion has a nucleus of charge +Ze and a single electron outside this nucleus. The energy levels of the ion are -Z^2RH/n^, where Z is the atomic number
okay I saw the rydberg formula but how am i or what numbers am i supposed to plug where
I don't get that. Your work LOOKS ok to me except your answer is 1/lambda. You need to take the reciprocal, right?
so i got 1/4797297.498
Plug that into your calculator; i.e., divide 1 by that large number and the answer will be the wavelength. That's what you are trying to find.
One note.
Since you are carrying all the other numbers to such high precision, I would suggest you carry the 0.0486 you obtained to the same number of significant figures.
That's what I have. Usually the answer is expressed either in Angstrom units or in nanometers. Your answer is 208.5 nm or 2085 Angstroms. Again, if you want to carry that many places for RH, then you are justified in more placed in the answer. I don't know how your professor wants it done. Many use 1.097 x 10^7 for RH.
The energy levels are
En = -Z^2RH/n^2
You left out the 2.
In your case, calculate En for Z = 3, with n = 4 and again for Z = 3 and n=3. Call those energy numbers E4 and E3.
E4 - E3 is the wave number (1/wavelength) of the photon emitted.
Invert that to get the wavelength
DrWLS showed you what to do.
1/lambda = RHZ2*[(1/N12)-(1/N22].
RH = 1.0967758341 x 10^7
Z = 3
N1 = 3
N2 = 4
Solve for lambda.
okay so it should look like this
1/lambda= 1.0967758341 x 10^7(3^2)*[(1/3^2)-(1/4^2)]
1.0967758341 x 10^7(9)(.0486)
i got 4798394.274
so I did 1 divided by 4797297.498
and got 2.085 X 10^-7
To find the wavelength of the transition from n=4 to n=3 for Li2+, you can use the Rydberg formula. The Rydberg formula states that the wavelength (λ) of the emitted or absorbed light during a transition is given by:
1/λ = RZ^2[(1/n_f^2) - (1/n_i^2)]
Where:
- λ is the wavelength of the light
- R is the Rydberg constant (approximately 1.097 × 10^7 m^-1)
- Z is the atomic number (in this case, it is +3 for Li2+)
- n_f is the final energy level (n=3 in this case)
- n_i is the initial energy level (n=4 in this case)
Plugging in the values into the formula:
1/λ = (1.097 × 10^7 m^-1) × (3^2) × [(1/3^2) - (1/4^2)]
Simplifying the expression:
1/λ = (1.097 × 10^7 m^-1) × 9 × [(1/9) - (1/16)]
= (1.097 × 10^7 m^-1) × 9 × [(16 - 9)/(9 × 16)]
= (1.097 × 10^7 m^-1) × 9 × (7/144)
Now, you can solve for the wavelength by taking the reciprocal of both sides:
λ = 1/[(1.097 × 10^7 m^-1) × 9 × (7/144)]
= 144/(1.097 × 10^7 m^-1 × 9 × 7/144)
= 144/(1.097 × 10^7 × 9 × 7) m
Calculating this expression will give you the wavelength of the transition from n=4 to n=3 for Li2+.
To determine the region of the spectrum in which this emission occurs, you can use the general guidelines for different regions:
- If the calculated wavelength is in the range of approximately 400-700 nm, it falls in the visible light region.
- If it is below 400 nm, it falls in the ultraviolet (UV) range.
- If it is above 700 nm, it falls in the infrared (IR) range.
Evaluate the calculated wavelength and determine in which region the emission occurs based on the above criteria.