Respuesta :
energy provided=13.4 eV
energy released=0.7 eV
energy absorbed = 12.7eV
initial energy= -13.6eV [ground state]
final energy = -13.6+12.7 = -0.9 eV
This energy corresponds to n=4 in hydrogen atom
energy released=0.7 eV
energy absorbed = 12.7eV
initial energy= -13.6eV [ground state]
final energy = -13.6+12.7 = -0.9 eV
This energy corresponds to n=4 in hydrogen atom
Answer:
n=5
Explanation:
We use the Rydberg formula to solve this problem:
1 / λ =[tex]R_{H}*(\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}} )[/tex]
Where λ is the wavelength of absorption/emission, RH is a constant (1.097*10⁷m⁻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 = 1.097*10⁷m⁻¹ * [tex](\frac{1}{1^{2}}-\frac{1}{n_{2}^{2}} )[/tex]
0.988 = [tex]1-\frac{1}{n_{2}^{2}} [/tex]
[tex]\frac{1}{n_{2}^{2}}=0.012\\[/tex]
n_{2}^{2}=83.333\\n_{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:
[tex]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\\n_{2}^{2}=26.50\\n_{2}=5.14[/tex]
So the final state of the hydrogen atom is n=5