Which two options are forms of kinetic energy?
1 answer:
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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
= -k e² /2a₀ 1/n²
For a transition between two states we have
-
= -k e² /2a₀ (1/
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 / ² - 1 / no₀²)
1 / λ = - k e² /2a₀ hc (1 / ² -1 / n₀²)
Where the constants are the Rydberg constant = 1.097 10⁷ m⁻¹
1 / λ = (1 / n₀² - 1 / nf²)
Now we can substitute the given values
Part A
Initial state n₀ = 1 to the final state = 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 = 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
The energy carried by a single photon is given by
where h is the Planck's constant and f is the frequency of the photon.
The photon of our exercise has a frequency of
, therefore its energy is
where h is the Planck's constant and f is the frequency of the photon.
The photon of our exercise has a frequency of