Question

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

Answer 1

Velocity of electrons
`(1)/(275) xx `velocity of light
`= (1)/(275) xx 3 xx 10^(10) = 1.09 xx10^(8) cm s^(-1)`
Since `u_(n) = (2pi e^(2))/(mh)`
`:. 1.09 xx 10^(8) = ( 2 xx 3.14 xx (4.803 xx 10^(-10))^(2))/(6.626 xx 10^(-27) xx n)`
` :. n = 20.06 xx 10^(-1) = 2 ` (an interger value)
Also when the electron jumps from `n + 1,` i.e. third to ground state
`bar v = (1)/(lambda) = R_(H) [(1)/(1^(2)) - (1)/(3^(2))]`
`= 109678 [(1)/(1) - (1)/(9)] = 9.75 xx 10^(4) cm^(-1)`