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 emission spectral lines
`N = (n(n-1))/(2) therefore 3 = (n_(1)(n_(1)-1))/(2)` in first case.
or `n_(1)^(2) - n_(1) -6 = 0` or `(n_(1)-3)(n_(1) +2) = 0`
Take positive root.
`therefore n_(1) = 3` ...(i)
Again, ` 6 = (n_(2)(n_(2)-1))/(2)` in second case.
or `n_(2)^(2) - n_(2) -12 =0` or `(n_(2)-4)(n_(2) + 3) = 0`
Take positive root, or `n_(2) = 4`
Now velocity of electron ` v = (2piKZe^(2))/(nh)`
`therefore (v_(1))/(v_(2)) = (n_(2))/(n_(1)) = 4/3`