The energy e of the electron in a hydrogen atom can be calculated from the bohr formula: =e−ryn2 in this equation ry stands for the rydberg energy, and n stands for the principal quantum number of the orbital that holds the electron. (you can find the value of the rydberg energy using the data button on the aleks toolbar.) calculate the wavelength of the line in the emission line spectrum of hydrogen caused by the transition of the electron from an orbital with =n10 to an orbital with =n8. round your answer to 3 significant digits.

  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

that would be because of both gravity and friction. so     c

490 n

f=125n a=2.5m/s2