Electron Speed In Bohr Orbits

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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?

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Within the Bohr model the orbital speed of the electron is given by $v_n = \dfrac{v_1}{n}$, where $v_1 = 2.18	imes10^6$ 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, $v_3 = \dfrac{2.18	imes10^6}{3} \approx 7.27	imes10^5$ m/s. The value $2.18	imes10^6$ m/s is the ground-state speed itself and ignores that higher orbits are slower. The value $1.09	imes10^6$ m/s wrongly divides by 2 rather than by $n = 3$. The value $6.54	imes10^6$ 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.

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