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 `(mvr) = n (h)/(2pi) = 4.22 xx 10^(-34) kg m^(2) s^(-1)`
`:. n = 4.22 xx 10^(-34) xx (2pi)/(h) = (2 xx 4.22 xx 10^(-34) xx 3.14)/(6.626 xx 10^(-34)) = 4` (Given)
When the electron jumps from n = 4 to n = 3, the wavelength of the spectral line can be calculated as follows:
`(1)/(lamda) = R_(H) ((1)/(n_(1)^(2)) - (1)/(n_(2)^(2))) = 109, 677 cm^(-1) ((1)/(3^(2)) - (1)/(4^(2))) = 109677 xx ((1)/(9) - (1)/(16)) = 109677 xx (7)/(144) cm^(-1)`
or `lamda = (144)/(109677 xx 7) cm = 1.88 xx 10^(-4) cm`