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:
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:
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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.
Updated On: Jun 8, 2026
- 3:4
- 2:3
- 3:2
- 4:3
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The Correct Option is A
Solution and Explanation
Concept:
The total number of possible emission spectral transition lines from a principal quantum state level \( n \) down to the ground state is given by the combination formula:
\[
N = \frac{n(n-1)}{2}
\]
According to Bohr's atomic model, the orbital velocity \( v \) of an electron in any given shell is inversely proportional to its principal quantum number \( n \):
\[
v \propto \frac{1}{n} \implies \frac{v_1}{v_2} = \frac{ndouble}{n1}
\]
Step 1: Finding the orbit level \( n_1 \) for Case 1.
Given total transitions \( N_1 = 6 \): \[ \frac{n_1(n_1-1)}{2} = 6 \implies n_1(n_1-1) = 12 \implies n_1 = 4 \]
Step 2: Finding the orbit level \( n_2 \) for Case 2.
Given total transitions \( N_2 = 3 \): \[ \frac{n_2(n_2-1)}{2} = 3 \implies n_2(n_2-1) = 6 \implies n_2 = 3 \]
Step 3: Calculating the orbital electron velocity ratio.
Using the inverse relationship between velocity and quantum level number: \[ \frac{v_1}{v_2} = \frac{n_2}{n_1} = \frac{3}{4} \]
Step 1: Finding the orbit level \( n_1 \) for Case 1.
Given total transitions \( N_1 = 6 \): \[ \frac{n_1(n_1-1)}{2} = 6 \implies n_1(n_1-1) = 12 \implies n_1 = 4 \]
Step 2: Finding the orbit level \( n_2 \) for Case 2.
Given total transitions \( N_2 = 3 \): \[ \frac{n_2(n_2-1)}{2} = 3 \implies n_2(n_2-1) = 6 \implies n_2 = 3 \]
Step 3: Calculating the orbital electron velocity ratio.
Using the inverse relationship between velocity and quantum level number: \[ \frac{v_1}{v_2} = \frac{n_2}{n_1} = \frac{3}{4} \]
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