The angular momentum of an electron in Bohr's orbit of hydrogen atom is 4.22 xx 10^(-34) kg m^(2) s^(-1). Calculate the wavelength of the spectral line when the electron falls from this level to the next lower level.

The angular momentum of an electron in Bohr's orbit of hydrogen atom i

1.	The angular momentum of an electron in Bohr's orbit of hydrogen atom is 4.22 xx 10^(-34) kg m^(2) s^(-1). Calculate the wavelength of the spectral line when the electron falls from this level to the next lower level.
Answer» Solution :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`