The angular momentum of an electron in Bohr's orbit of H-atom is `3.02 xx 10^(-34) kg m^2 s^(-1)`. Calculate the wavelength of the spectral line emitted when the electron jumps this level to the next lower level.

Explanation

Angular momentum `(mvr)`
`=n h/(2pi) = 3.02 xx 10^(-34) kg m^2 s^(-1)`
`n = 3.02 xx 10^(-34) xx (2pi)/(h)`
`= (3.02 xx 10^(-34) xx 2 xx 3.14)/(6.3 xx 10^(-34)) = 3`.
When the electron jumps from `n = 3 ` to `n = 2` , the wavelength of spectral line,
`1/lambda = 109677 (1/(n_2^2) - 1/(n_1^2))`
`= 109677 (1/(2^2) - 1/(3^2))`
`1/lambda = 109677 xx 5/36`
or `lambda = 36/(5 xx 109677)`
`= 6.56 xx 10^(-5) cm " or " 656 nm`.