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 speed of the electrons in the orbits are in the ratio

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

Answer 1

Answer: (A)

Number of emitted spectral lines $N=\frac{n(n-1)}{2}$ 1st case $N=3$ $\therefore$ $3=\frac{n_1(n_1-1)}{2}$ $\Rightarrow$ $n_1^2-n_1-6=0$ $\Rightarrow$ $(n_1-3)(n_1+2)=0$ $n_1=3,n_1=-2$ Negative value of n^ is not possible $\therefore$ $n_1=3$ 2nd case $N=6$ Again, $6=\frac{n_2(n_1-1)}{2}$ $\Rightarrow$ $n_2^2-n_2-12=0$ $\Rightarrow$ $(n_2-4)(n_2+3)=0$ $n_2=4,n_2=-3$ Again, as $n_2$ is always positive $\therefore$ $n_2=4$ Velocity of electron $v=\frac{Z{e}^2}{2{\epsilon }_{0}hn}$ $\therefore$ $\frac{v_1}{v_2}=\frac{n_2}{n_1}$ $\Rightarrow$ $\frac{v_1}{v_2}=\frac{4}{3}$