Question

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 speeds of the electrons in the two orbits are in the ratio

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

Answer 1

Answer: (D)

Number of emitted spectral lines
$N=\frac{n(n-1)}{2}$
1st case $N=3$
$\therefore 3 =\frac{n_{1}\left(n_{1}-1\right)}{2} $
$\Rightarrow n_{1}^{2}-n_{1}-6 =0$
$\Rightarrow \left(n_{1}-3\right)\left(n_{1}+2\right) =0 $
$ n_{1}=3, n_{1} =-2$
Negative value of $n_{1}$ is not possible
$\therefore n_{1}=3$
2nd case $ N=6$
Again, $6=\frac{n_{2}\left(n_{2}-1\right)}{2}$
$\Rightarrow n_{2}^{2}-n_{2}-12=0$
$\rightarrow\left(n_{2}-4\right)\left(n_{2}+3\right)=0$
$n_{2}=4, n_{2}=-3$
Again, as $n_{2}$ is always positive
$\therefore n_{2}=4$
Velocity of electron $v=\frac{Z e}{2 \varepsilon_{0} h n}$
$\therefore \frac{v_{1}}{v_{2}}=\frac{n_{2}}{n_{1}}$
$\Rightarrow \frac{v_{1}}{v_{2}}=\frac{4}{3}$