Question

What is the wavelength of photon emitted during a transition from n=5 to n=2 state in Be+2 ion?

Answer 1

What are you able to do on your own? I'll help ya out in explaining anything you need, I think a good place to start is do you have any equations that tell you the energy of a transition between two different states? It should look something like:

\[E \approx \frac{1}{n_1^2} - \frac{1}{n_2^2}\]

from there you'll relate this through two other formulas, \(E= h \nu\) and \(c = \lambda \nu\). Try to explain what you understand and what you don't so I can help you.

Answer 2

Okay. I do know the equation.
\[\frac{ 1 }{ \lambda } = R \left( \frac{ 1 }{ n _{1}^2 } -\frac{ 1 }{ n _{2}^2 } \right) \]
But this applies to hydrogen atom only. I don't understand how do I use this for Be+2 ion?

Answer 3

Yeah this should only apply to Hydrogen-like atoms like Be3+ not Be2+, but maybe this is just as good, I'm not sure if this will apply. At any rate I wasn't sure so I looked it up, it appears to depend on the square of the charge of the nucleus, \(Z^2\) and possibly the reduced mass of the system depending on how in depth you want to get:

\[\frac{1}{\lambda} = R Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)\]

Here's where I got that from, might be worth checking out:

https://en.wikipedia.org/wiki/Rydberg_formula#Rydberg_formula_for_any_hydrogen-like_element

Answer 4

Okay. Thank you so much. :)