Electrons in a certain energy level `n=n_(1)` can emit 3 spectral lines. When they are in another energy level, `n=n_(2)`, they can emit 6 spectral lines. The orbital speed of the electrons in the two orbits are in the ratio

Options

A. `4:3`
B. `3:4`
C. `2:1`
D. `1:2`

Answer

A

Explanation

Number of emitted spectral lines,
`N=(n(n-1))/(2)`
Case I `N=3`
`therefore 3=(n_(1)(n_(1)-1))/(2)`
`implies n_(1)^(2)-n_(1)-6=0 implies (n_(1)-3)(n_(1)+2)=0`
`n_(1)=3,n_(1)=-2`
Negative value of `n_(1)` is not possible.
`therefore n_(1)=3`
Case II `N=6`
Again, `6=(n_(2)(n_(2)-1))/(2)`
`implies n_(2)^(2)-n_(2)-12=0`
`implies (n_(2)-4)(n_(2)+3)=0`
`n_(2)=4,n_(2)=-3`
Again, as `n_(2)` is always positive
`therefore n_(2)=4`
Velocity of electron, `v=(Ze^(2))/(2epsi_(0)hn)`
`therefore (v_(1))/(v_(2))=(n_(2))/(n_(1)) implies (v_(1))/(v_(2))=(4)/(3)`