The angular momentum of electron in a Bohr's orbit of H atom is `4.2178 xx 10^(-34) kg m^(2)s^(-1)`. Calculate the wavelength of the spectral line when the electrton falls from this level to the next lower level.

Explanation

Angular momentum of an electron in a Bohr.s orbit of H-atom
`mvr = (nh)/(2 pi)`
`4.218 xx 10^(-34) kg m^(2)s^(-1) = (n xx 6.626 xx 10^(-34) kg m^2 s^(-2))/(2 xx 22/7)`
or `" " n = (4.218 xx 10^(-34) xx 2 xx 22)/(6.626 xx 10^(-34) xx 7) = 4`.
Now, `" " barv = 1/lambda = 109678 (1/(n_1^2) - 1/(n_2^2)) cm^(-1)`
The spectral line when electron falls from `4^(th)` level to `3rd` level so that
`n_2 = 4, n_1 = 3`
`1/(lambda) = 109678 (1/(3^2) - 1/(4^2)) cm^(-1)`
`1/lambda = 109678 ((16 - 9)/(9 xx 16)) cm^(-1)`
`1/lambda = 109678 xx 7/(9 xx 16) cm^(-1)`
`:. lambda = (9 xx 16)/(109678 xx 7) = 1.876 xx 10^(-4) cm`