An electron in a Bohr orbit of hydrogen atom with the quantum number `N_(2)` has an angular momentum `4.2176 xx 10^(-34) kg m^(2) s^(-1)`. If the electron drops from this level to the next lower level , the wavelength of this lines is

Options

A. `18 nm`
B. `187.6 pm`
C. `1876 Å`
D. `1.876 xx 10^(4) Å`

Answer

D

Explanation

Angular momentum, `L = 4.2176 xx 10^(-34) = (n_(2) h)/(2 pi)`
`implies n_(2) = 4`
For the transtion from `n_(2) = 4 to n_(1) = 3`, the wavelength spectral line`= lambda`
`(1)/(lambda) = (13.6)/(hc) ((1)/(3^(2)) - (1)/(4^(2)))`
`= (13.6 eV)/(1240 eV nm) ((7)/(9 xx 16))`
`lambda = (1240 xx 144)/(13.6 xx7) = 1876 nm = 18760 Å`
`= 1.876 xx 10^(4) Å`