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

Question

Dr. Abubacker · Accepted Answer

(a) Number of emission spectral lines
$N = \frac{n(n - 1)}{2}$
$	herefore 3 = \frac{n_{1}(n_{1} - 1)}{2}$, in first case
Or $n_{1}^{2} - n_{1} - 6 = 0$ or $\left( n_{1} - 3 ight)\left( n_{1} + 2 ight) = 0$
Take positive root
$	herefore n_{1} = 3$
Again, $6 = \frac{n_{2}(n_{2} - 1)}{2}$, in second case
Or $n_{2}^{2} - n_{2} - 12 = 0$ or $\left( n_{2} - 4 ight)\left( n_{2} + 3 ight) = 0$
Take positive root, or $n_{2} = 4$
Now velocity of electron $v = \frac{2\pi KZe^{2}}{nh}$
$	herefore\frac{v_{1}}{v_{2}} = \frac{n_{2}}{n_{1}} = \frac{4}{3}$

Q. 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

Explanation

Reviewed by Dr. Abubacker

Want more practice?