The angular momentum of an electron present in a Bohr orbit is `3.1655xx10^(-34)kg*m^(2)*s^(-1)`. What will be the wavelength of the spectral line produced when the electron jumps to the adjacent orbit with lowest radius?

Explanation

The angular momentum of electron revolving in the n-th Bohr orbit is given by, `mvr=(nh)/(2pi)`
`therefore 3.1655xx10^(-34)=(nxx6.626xx10^(-34))/(2xx3.14) or, n=3`
Let, `lamda` be the wave length of the spectral line produced when the electron jumps from n=3 to n=2
`therefore overline(v)=(1)/(lamda)=R_(H)((1)/(n_(1)^(2))-(1)/(n_(2)^(2)))=109678((1)/(2^(2))-(1)/(3^(2)))`
`=109678xx(5)/(36)`
`therefore lamda=(36)/(5xx109678)=6.564xx10^(-5)cm`