Question

The velocity of electron in a certain Bohr orbit of H bears the ratio `1: 275 ` to the velocity of light M
a. What is the quantum number (n) of orbit ?
b. Calculate the wavelength of the radiation emitted whemn the electron jumps from `(n + 1)` state to the ground state `(R = 1.0987 xx 10^(5) cm^(-1))`

Answer 1

Given V = `("Velocity of light")/(275) = (3 xx 10^(8))/(275)`
`V_(n) = 2.165 xx 10^(6) (Z)/(n)`
(a) `2.165xx10^(6)((1))/n = (3 xx 10^(8))/(275) Rightarrow n = 2 `
b. `(n+ 1)` i.e. `3 rarr 1 ("ground state")`
`bar v = 1.0987 xx 10^(7)((1)/(1^(2)) - (1)/(2^(2)))`
`= 9.76 xx 10^(6) m^(-1) = 9.76 xx 10^(4)cm^(-1)`