An electron in a Bohr orbit of hydrogen atom in quantum level `n_(2)` has an angular momuntum `4.2176 xx 10^(-34) kg m^(2)s^(-1)` .If this electron drops from this level to the next level , find the wavelength of this spectral line (Gives `R_(H) = 10979 cm^(-1))`

Explanation

According to Bohr's theory
`mvr = (nh)/(2pi)`
`mvr = 4.2176 xx 10^(-34) kg m^(2)s^(-1)= nh//2pi`
`n = (2 xx 3.14 xx 4.2176 xx 10^(-34) kg m^(2)s^(-1))/((6.626 xx 10^(-34)J s))`
When an electron falls from `n = 4` to` n = 3` in a hydrogen atom , the wavelength emitted is calcularated as
`(1)/(lambda) = R_(H)[(1)/(n_(1)^(2)) - (1)/(n_(2)^(2))] = 109679[(1)/(3^(2)) - (1)/(4^(2))]`
`lambda = 1.8756 xx 10^(-4) cm^(-1)`
`= 1.8756 xx 10^(-6)m^(-1)`
`= 1.8756 xx 10^(10)m^(-1)`
`= 18756 Å`