The image above was taken by the spacecraft Messenger as it flew by the planet Mercury. The terrain shown in this image is typic
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Answer:
λ = 162 10⁻⁷ m
Explanation:
Bohr's model for the hydrogen atom gives energy by the equation
= - k²e² / 2m (1 / n²)
Where k is the Coulomb constant, e and m the charge and mass of the electron respectively and n is an integer
The Planck equation
E = h f
The speed of light is
c = λ f
E = h c /λ
For a transition between two states we have
-
= - k²e² / 2m (1 /
² -1 /
²)
h c / λ = -k² e² / 2m (1 / ² - 1/
²)
1 / λ = (- k² e² / 2m h c) (1 / ² - 1/
²)
The Rydberg constant with a value of 1,097 107 m-1 is the result of the constant in parentheses
Let's calculate the emission of the transition
1 /λ = 1.097 10⁷ (1/10² - 1/8²)
1 / λ = 1.097 10⁷ (0.01 - 0.015625)
1 /λ = 0.006170625 10⁷
λ = 162 10⁻⁷ m