contestada

Part A: Determine the wavelength of photons that can be emitted
from of the n=4 state of a hydrogen atom to the n=1 state. Give
answer in nm .
Part B: determine the wavelength of photons that can it be
emitted from of the n=4 state of a hydrogen atom to the n=2
state. Give answer in nm .
Part C: Determine the wavelength of photons that can be
emitted from of the n=4 state of a hydrogen atom to the n=3 state.
Give answer in nm .

Respuesta :

Answer:

A  λ = 97.23 nm

, B)   λ = 486.2 nm

, C)  λ = 53326 nm

Explanation:

With that problem let's use the Bohr model equation for the hydrogen atom

          [tex]E_{n}[/tex] = -k e² /2a₀  1/n²

For a transition between two states we have

          [tex]E_{nf}[/tex] -  [tex]E_{no}[/tex] = -k e² /2a₀ (1/  [tex]n_{f}[/tex]² - 1 / n₀²)

Now this energy is given by the Planck equation

         E = h f

And the speed of light is

         c = λ f

Let's replace

      h c / λ = - k e² /2a₀ (1 / [tex]n_{f}[/tex]² - 1 / no₀²)

      1 /  λ = - k e² /2a₀ hc (1 / [tex]n_{f}[/tex]² -1 / n₀²)

Where the constants are the Rydberg constant [tex]R_{H}[/tex] = 1.097 10⁷ m⁻¹

        1 /  λ = [tex]R_{H}[/tex] (1 / n₀² - 1 / nf²)

Now we can substitute the given values

Part A

 Initial state n₀ = 1 to the final state [tex]n_{f}[/tex] = 4

        1 /  λ = 1.097 10⁷ (1/1 - 1/4²)

         1 /  λ = 1.0284 10⁷ m⁻¹

          λ = 9.723 10⁻⁸ m

We reduce to nm

         λ = 9.723 10⁻⁸ m (10⁹ nm / 1m)

         λ = 97.23 nm

Part B

Initial state n₀ = 2 final state[tex]n_{f}[/tex] = 4

       1 /  λ = 1.097 10⁷ (1/2² - 1/4²)

       1 /  λ = 0.2056 10⁻⁷ m

        λ = 486.2 nm

Part C

Initial state n₀ = 3

      1 /  λ = 1,097 10⁷ (1/3² - 1/4²)

       1 /  λ = 5.3326 10⁵ m⁻¹

        λ = 5.3326 10-5 m

        λ = 53326 nm

Otras preguntas