Question

An electron is moving in an orbit of hydrogen atom in which there can be a maximum of six transitions. Another electron is moving in another orbit of hydrogen atom in which there can be a maximum of three transitions. The ratio of the velocity of electrons in these two orbits is:

• A. 3:4
• B. 2:3
• C. 3:2
• D. 4:3

Hint:

Higher orbital index numbers represent outer energy shells where the electron experiences a weaker electrostatic pull from the nucleus, resulting in a slower orbital velocity.

Answer 1

The Correct Option is A

Step 1: Connect transitions to the orbit number.
The number of spectral lines an electron can give while falling from level $n$ is: \[ N = \frac{n(n-1)}{2} \] We use this to find the orbit number in each case.
Step 2: Find the first orbit number.
Given $N_1 = 6$: \[ \frac{n_1(n_1-1)}{2} = 6 \;\Rightarrow\; n_1(n_1-1) = 12 \;\Rightarrow\; n_1 = 4 \]
Step 3: Find the second orbit number.
Given $N_2 = 3$: \[ \frac{n_2(n_2-1)}{2} = 3 \;\Rightarrow\; n_2(n_2-1) = 6 \;\Rightarrow\; n_2 = 3 \]
Step 4: Recall how speed depends on the orbit.
In the Bohr model the electron speed gets smaller in higher orbits: \[ v \propto \frac{1}{n} \]
Step 5: Write the speed ratio.
Because speed is inversely proportional to $n$, the ratio flips: \[ \frac{v_1}{v_2} = \frac{n_2}{n_1} = \frac{3}{4} \]
Step 6: State the answer.
\[ \boxed{v_1 : v_2 = 3 : 4} \]