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, $6=\frac{n_2\left(n_2-1\right)}{2}$
$\Rightarrow n_2^2-n_2-12=0$
$\to \left(n_2-4\right)\left(n_2+3\right)=0$
$n_2=4,n_2=-3$
Again, as $n_2$ is always positive
$\therefore n_2=4$
Velocity of electron $v=\frac{Ze}{2{\epsilon }_{0}hn}$
$\therefore \frac{v_1}{v_2}=\frac{n_2}{n_1}$
$\Rightarrow \frac{v_1}{v_2}=\frac{4}{3}$