Question

A transition of an electron between two different energy levels produced a series of line spectrum . calculate the wavelength of (β) line in Paschen series. wavelength in nanometer?

Answer 1

I found: `lambda=1282 nm`

Explanation:

Considering your `beta` line in the Paschen series as a transition from n = 5 to n = 3:
http://www.daviddarling.info/encyclopedia/L/Lyman_series.html
I would use Rydberg Formula:
`1/lambda=R(1/n_(low)^2-1/n_(up)^2)`
Where:
`lambda` is the wavelength;
`R` is Reydberg Constant (`=1.097373xx10^7 m^-1`);
`n_(low) and n_(up)` are the two quantum numbers caracterizing the transition.
so you get:
`1/lambda=1.097373xx10^7(1/3^2-1/5^2)=7.8xx10^5m^-1`
`lambda=1282 nm`