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 wave number 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

Correct Answer - `overline(v)=9.75xx10^(4)cm^(-1)`.
(a). `(v)/(c)=(1)/(275)` or `v=(3xx10^(10))/(275)=1.09xx10^(8)cm" "sec^(-1)`
`v=(nh)/(2pimr)=(nh)/(2pimxx0.529xx10^(-8)xxn^(2))`
or `n=(h)/(2pimxx0.529xx10^(-8)xxv)`
`=(6.625xx10^(-27))/(2xx3.14xx9.1xx10^(-28)xx0.529xx10^(-8)xx1.09xx10^(8))`
`=2`
(b). Thus, `n+1=2+1=3`. the electron jumps from 3rd orbit to 1st orbit.