Question

The energy E of the electron in a hydrogen atom can be calculated from the Bohr formula: E = R_y/n^2 In this equation R_y 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 n = 11 to an orbital with n = 10. Round your answer to 3 significant digits.

Answer 1

The wavelength of the line in the emission line spectrum of hydrogen caused by the transition of the electron for the given energy levels is
`5.23* 10^(-5) m`

Step-by-step explanation:

Given :

The energy E of the electron in a hydrogen atom can be calculated from the Bohr formula:

`E=(R_y)/(n^2)`

`R_y=2.18* 10^(-18) J` = Rydberg energy

n = principal quantum number of the orbital

Energy of 11th orbit =
`E_(11)`

`E_(11)=(2.18* 10^(-18) J)/(11^2)=1.80* 10^(-20) J`

Energy of 10th orbit =
`E_(10)`

`E_(10)=(2.18* 10^(-18) J)/(10^2)=2.18* 10^(-20) J`

Energy difference between both the levels will corresponds to the energy of the wavelength of the line which can be calculated by using Planck's equation.

`E'=E_(10)-E_(11)=2.18* 10^(-20) J-1.80* 10^(-20) J`

`=E'=0.38* 10^(-20) J`

`\lambda =(hc)/(E')` (Planck's' equation)

`\lambda = (6.626* 10^(-34) Js* 3* 10^8 m/s)/(0.38* 10^(-20) J)`

`\lambda = 5.2310* 10^(-5) m\approx 5.23* 10^(-5) m`

The wavelength of the line in the emission line spectrum of hydrogen caused by the transition of the electron for the given energy levels is
`5.23* 10^(-5) m`