Electron Speed In Bohr Orbits
A hydrogen electron occupies the third Bohr orbit, where its orbital speed is smaller than in the ground state. If the ground-state orbital speed is 2.18 × 10^6 m/s, what is the electron's speed in this third orbit?
Select the correct option:
Solution
7.27×105m/s
Within the Bohr model the orbital speed of the electron is given by vn=nv1, where v1=2.18×106 m/s is the ground-state speed; the speed therefore falls inversely with the principal quantum number. This inverse dependence comes from combining angular-momentum quantisation with the Coulomb-centripetal balance. For the third orbit, v3=32.18×106≈7.27×105 m/s. The value 2.18×106 m/s is the ground-state speed itself and ignores that higher orbits are slower. The value 1.09×106 m/s wrongly divides by 2 rather than by n=3. The value 6.54×106 m/s multiplies by 3 instead of dividing, reversing the correct trend. The orbital speed in the ground state is about 1/137 of the speed of light, the ratio known as the fine-structure constant, which justifies treating the hydrogen electron non-relativistically. As n rises the electron both slows down and shifts to a larger orbit, so its kinetic energy falls steadily. This follows the standard NCERT Bohr-orbit velocity relation. A plausibility check confirms the speed remains far below the speed of light and is smaller than the ground-state value, consistent with electrons in outer orbits moving more slowly.
🔒 Solution Hidden from View
Submit your answer to unlock the detailed step-by-step solution.
About This Question
- Subject
- physics
- Chapter
- atoms and nuclei
- Topic
- electron speed in bohr orbits
- Difficulty
- Medium
- Year
- 2025
Solution
Correct Answer:
7.27×105m/s
Within the Bohr model the orbital speed of the electron is given by vn=nv1, where v1=2.18×106 m/s is the ground-state speed; the speed therefore falls inversely with the principal quantum number. This inverse dependence comes from combining angular-momentum quantisation with the Coulomb-centripetal balance. For the third orbit, v3=32.18×106≈7.27×105 m/s. The value 2.18×106 m/s is the ground-state speed itself and ignores that higher orbits are slower. The value 1.09×106 m/s wrongly divides by 2 rather than by n=3. The value 6.54×106 m/s multiplies by 3 instead of dividing, reversing the correct trend. The orbital speed in the ground state is about 1/137 of the speed of light, the ratio known as the fine-structure constant, which justifies treating the hydrogen electron non-relativistically. As n rises the electron both slows down and shifts to a larger orbit, so its kinetic energy falls steadily. This follows the standard NCERT Bohr-orbit velocity relation. A plausibility check confirms the speed remains far below the speed of light and is smaller than the ground-state value, consistent with electrons in outer orbits moving more slowly.
This medium difficulty physics question is from the chapter atoms and nuclei, covering the topic of electron speed in bohr orbits. It appeared in the 2025 exam.
Looking for more practice? Explore all physics questions or browse atoms and nuclei questions on RankGuru.