Answer: 3
Explanation:
You want to use the Rydberg equation which is given by
[tex]E = -R_{H}(\frac{1}{n_{final}^2} - \frac{1}{n_{initial}^2})[/tex]
where [tex]R_H = 2.18 \, \cdot 10^{-18} J[/tex].
We are however given the wavelength instead of energy, and so we need to represent E in terms of its wavelength, and so we use [tex]E = \frac{hc}{\lambda}[/tex]. Remember that λ must be in meters in this formula. So we can divide 1284 nm by [tex]10^9[/tex] to get the wavelength in meters. This gives [tex]\lambda = 1.284 \, \cdot 10^{-6}[/tex] meters. Putting this all together we have
[tex]\frac{hc}{\lambda} = -R_{H}(\frac{1}{n_{final}^2} - \frac{1}{n_{initial}^2})[/tex]
where
[tex]h = 6.626 \cdot 10^{-34} J\cdot s\\c = 2.998 \cdot 10^{8}\, m/s\\\lambda = 1.284 \cdot 10^{-6}\, m\\R_H = 2.18 \, \cdot 10^{-18} J\\n_{initial} = 5\\n_{final} = \, ?[/tex]
If we plug in everything we should get roughly [tex]n_f = 3.00194[/tex] but remember that orbitals must be a whole number and so you would round to the closest integer which gives [tex]n_f = 3.[/tex]
