Answer:
n=5
Explanation:
We use the Rydberg formula to solve this problem:
1 / λ =\(R_{H}*(\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}} )\)
Where λ is the wavelength of absorption/emission, RH is a constant (1.09710⁷m⁻¹), n₁ is the state of minor energy and n₂ is the state with higher energy.
- First, for the description for the absorption we have λ=92.3 10⁻⁹m, and n₁ = 1, so we solve for n₂:
1 / 92.3 10⁻⁹m = 1.09710⁷m⁻¹ \((\frac{1}{1^{2}}-\frac{1}{n_{2}^{2}} )\)
0.988 = \(1-\frac{1}{n_{2}^{2}}\)
\(\frac{1}{n_{2}^{2}}=0.012\\\)
n_{2}^{2}=83.333\
_{2}=9.12[/tex]
So after absorbing the wavelength of 92.30 nm the state of the hydrogen atom is n=9
- Now for the emission, we have λ=1820 10⁻⁹m = 1.82*10⁻⁶m, and n₂=9, so we solve for n₁:
\(1 / 1.82*10^{-6}m = 1.097*10^{7}m*(\frac{1}{n_{2}^{2}}-\frac{1}{9^{2}} )\\5.009*10^{-2}=\frac{1}{n_{2}^{2}}-\frac{1}{81}\\\frac{1}{n_{2}^{2}}=0.0377\
_{2}^{2}=26.50\
_{2}=5.14\)
So the final state of the hydrogen atom is n=5
