Question

Calculate the wavelength of light (in nm) of the spectral line of Hydrogen where an electron falls from the 6th Bohr orbit to the 3rd Bohr orbit.

a) 540 nm
b) 2000 nm
c) 1090 nm
d) 1050 nm

Answer 1

The wavelength of light (in nm) of the spectral line of Hydrogen where an electron falls from the 6th Bohr orbit to the 3rd Bohr orbit is 1090nm

Step-by-step explanation:

We know that , the wavelength of the light is calculated by Rydberg's formula-

`(1)/(\pi) =R^2((1)/(n^2_1) -(1)/(n^2_2))`
`[n_2>n_1]`

Here , R = Rydberg's constant
`(1.097* 10^7 m^-^1)`

Z = atomic number (for hydrogen , Z= 1)

`n_2 =6 , n_1=3`

`\pi =` wavelength of light

Now , putting the values in the Rydberg's formula ,

`(1)/(\pi) =1.097*10^7m^-^1((1)/(3^2) -(1)/(6^2) )`

=
`1.097* 10^7m^-^1 ((4-1)/(36) )`

=
`1.097* 10^7m^-^1((3)/(36) )`

=
`1.097* 10^7m^-^1((1)/(12) )`

`(1)/(\pi)` =
`0.0914167* 10^7m^-^1`

`\pi=(1)/(0.0914167*10^7m^-^1)`

`\pi=10.9389*10^-^7\\\pi=1093.89*10^-^9m`

=1090nm

Hence , the wavelength of the light is 1090nm,, that is option D is correct.