Question

A ground state hydrogen atom absorbs a photon of light having a wavelength of 92.30 nm. It then gives off a photon having a wavelength of 1820 nm. What is the final state of the hydrogen atom?

Answer 1

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

Answer 2

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 (1.097*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 * $(\frac{1}{1^{2}}-\frac{1}{n_{2}^{2}}  )$

0.988 = $1-\frac{1}{n_{2}^{2}}  $

$\frac{1}{n_{2}^{2}}=0.012$

n_{2}^{2}=83.333\\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:

$1 / 1.82*10^{-6}m = 1.097*10^{7}m*(\frac{1}{n_{2}^{2}}-\frac{1}{9^{2}}  )\\5.009*10^{-2}=\frac{1}{n_{2}^{2}}-\frac{1}{81}\\\frac{1}{n_{2}^{2}}=0.0377\\n_{2}^{2}=26.50\\n_{2}=5.14$

So the final state of the hydrogen atom is n=5