Line Spectra series for Ionized Helium

[rhouli67]

Homework Statement

Adjacent wavelengths 251.1 nm and 238.6 nm are found in one series of transitions among the radiations emitted by ionized helium.

1.) Find the value of n₀ for this series:
Possible answers:
n0= 1, 2, 3, 4, 5

2.) Find the series limit:
Possible answers:
λ_limit= 91.13 nm, 205.0 nm, 364.5 nm, 569.5 nm

Homework Equations

λ=λ_limit*n²/(n²-n²₀) where (n=n₀+1, n₀+2,...)
Nuclear charge for Ionized Helium Z=2

The Attempt at a Solution

I think it need to find maximum energy of the helium atom and use the relation E =hc/λ => λ= hc/E to find the series limit, λ_limit, then plug in various values of n via trial and error until i get the right answer. I am however struggling trying to figure out how to find the maximum energy of the atom (assuming this is actually the way to approach the problem).

Thanks for your time.

 

[mark726]

Thank you for your post. I would like to offer some guidance in solving this problem.

Firstly, you are correct in your approach to finding the series limit. The series limit is the maximum wavelength (λlimit) in the series, which corresponds to the lowest energy transition. This means that the energy of the transition will be the difference between the energy levels of the atom in its ground state (n=1) and the energy level at the series limit (n=n0+1).

To find the series limit, you can use the Rydberg formula for the energy levels of the hydrogen atom:
E = -13.6 eV/n^2, where n is the principal quantum number. For helium, the nuclear charge is Z=2, so the energy levels will be E = -54.4 eV/n^2.

To find the maximum energy of the helium atom, we need to find the energy level at n=n0+1. We know that the two adjacent wavelengths are 251.1 nm and 238.6 nm, which correspond to transitions from n=3 to n=2 and from n=4 to n=3, respectively. This means that the series limit (n=n0+1) must be between n=3 and n=4.

Using the Rydberg formula, we can set up the following equations:
E1 = -54.4 eV/3^2 = -6.04 eV
E2 = -54.4 eV/4^2 = -3.40 eV

Since the energy of the transition is the difference between the energy levels, we can set up the equation:
E1 - E2 = 2.64 eV = hc/λlimit

Solving for λlimit, we get:
λlimit = 471.2 nm

Therefore, the series limit is 471.2 nm. Now, we can use the formula you provided to find the value of n0:
λlimit = λlimit*n2/(n2-n20)
471.2 nm = 471.2 nm*4/(4-n0)
Solving for n0, we get:
n0 = 3

So, the value of n0 for this series is 3.

I hope this helps. Let me know if you have any further questions or if you need clarification on any part of the solution