Velocity of Electron in nth Bohr Orbit: Formula and Derivation
Physics · Atoms · NEET
The speed of the electron in the nth Bohr orbit is v_n = (2 pi k Z e^2)/(n h) = e^2 Z / (2 epsilon_0 n h). For hydrogen (Z = 1) this gives v_n = 2.19 x 10^6 / n m/s, so the ground-state (n = 1) speed is about 2.19 x 10^6 m/s. Key idea: speed goes DOWN as n goes UP. Memory hook: "Velocity likes Z/n" - it grows with the nuclear charge Z and shrinks with the orbit number n.
The electron moves fastest in the innermost orbit (n = 1) and slows down in outer orbits because velocity is proportional to Z/n. Note radius grows as n^2 while speed falls as 1/n.
Your doubts, answered
Is the electron velocity proportional to n or to 1/n?
It is proportional to 1/n. The formula is v_n = 2.19x10^6 (Z/n) m/s, so as the orbit number n increases, the speed decreases. Students often confuse this with radius, which is proportional to n^2 (r grows with n). Speed drops, radius grows. So the outer orbits are bigger but slower.
What is the electron speed in the ground state of hydrogen?
For hydrogen Z = 1 and ground state n = 1, so v_1 = 2.19 x 10^6 m/s (about 2.2 x 10^6 m/s). This is roughly c/137, where c is the speed of light and 1/137 is the fine-structure constant alpha. So the electron moves at about 0.73% the speed of light.
Why does the electron move slower in higher orbits?
From Bohr quantisation, mvr = nh/2pi, so v = nh/(2pi m r). But the radius grows faster than n: r is proportional to n^2. Putting r ~ n^2 in gives v ~ n/n^2 = 1/n. The radius wins, so the net speed falls off as 1/n. Physically the electron is farther from the nucleus, feels a weaker pull, and orbits more slowly.
How does velocity depend on the atomic number Z?
Velocity is directly proportional to Z: v_n = 2.19x10^6 (Z/n) m/s. For He+ (Z = 2) the electron in n = 1 moves twice as fast as in hydrogen. A larger nuclear charge pulls the electron harder, so it needs a higher orbital speed to stay in a stable circular orbit.
How do you derive the velocity formula?
Set the Coulomb force equal to the centripetal force: kZe^2/r^2 = mv^2/r, giving mv^2 = kZe^2/r. Combine with Bohr quantisation mvr = nh/2pi. Solve the two together to eliminate r: v_n = 2pi k Z e^2/(n h) = Ze^2/(2 epsilon_0 n h). Numerically this is 2.19 x 10^6 (Z/n) m/s.
⚠️ The NEET trap ✗ Treating velocity like radius and writing v proportional to n^2, or thinking the electron speeds up in higher orbits. ✓ Velocity is proportional to Z/n (it decreases as n increases). Only the radius grows as n^2; speed and energy fall with n. 🧠 Radius grows, speed slows. r goes up as n^2, v goes down as 1/n.
Real NEET questions
NEET 2025
A particle of mass m is moving around the origin with a constant force F pulling it towards the origin. If Bohr's model is used to describe its motion, the radius of the nth orbit and the particle's speed v in the orbit depend on n as:
A · r proportional to n^(2/3) ; v proportional to n^(1/3) ✓
B · r proportional to n^(4/3) ; v proportional to n^(-1/3)
C · r proportional to n^(1/3) ; v proportional to n^(1/3)
D · r proportional to n^(1/3) ; v proportional to n^(2/3)
Solution: Step 1: The constant central force F provides the centripetal force, so F = mv^2/r, giving v^2 = Fr/m. Step 2: Apply Bohr quantisation mvr = nh/2pi, so v = nh/(2 pi m r). Step 3: Substitute into v^2 = Fr/m: (nh/2 pi m r)^2 = Fr/m. Step 4: n^2 h^2/(4 pi^2 m^2 r^2) = Fr/m, so r^3 = n^2 h^2/(4 pi^2 m F), giving r proportional to n^(2/3). Step 5: Then v^2 = Fr/m is proportional to r, which is proportional to n^(2/3), so v proportional to n^(1/3). Answer: option A. (Note: for the normal Coulomb 1/r^2 force you instead get v proportional to 1/n; this problem uses a constant force, so the scaling changes.)
ReNEET 2026
An electron is revolving in an excited state of a Hydrogen atom with velocity sqrt(25.6) x 10^5 m/s. The radius of the orbit is x x 10^-9 m. Find x. [m_e = 9x10^-31 kg, e = 1.6x10^-19 C, 1/(4 pi epsilon_0) = 9x10^9 N m^2 C^-2]
A · 4
B · 3
C · 2
D · 1 ✓
Solution: Step 1: The Coulomb force provides the centripetal force: k e^2/r^2 = m v^2/r, so r = k e^2/(m v^2). Step 2: Here v^2 = (sqrt(25.6) x 10^5)^2 = 25.6 x 10^10 m^2/s^2. Step 3: Substitute r = (9x10^9 x (1.6x10^-19)^2)/(9x10^-31 x 25.6x10^10). Step 4: Numerator = 9x10^9 x 2.56x10^-38 = 2.304x10^-28. Denominator = 9x10^-31 x 25.6x10^10 = 2.304x10^-19. Step 5: r = 2.304x10^-28 / 2.304x10^-19 = 1x10^-9 m. So x = 1. Answer: option D.
Solved Atoms NEET PYQs
Try the real previous-year questions from this chapter — each with the answer and a full solution.
What is the formula for velocity of electron in nth Bohr orbit?
v_n = (2 pi k Z e^2)/(n h) = Z e^2/(2 epsilon_0 n h). Numerically v_n = 2.19 x 10^6 (Z/n) m/s. It is directly proportional to Z and inversely proportional to n.
What is the velocity of electron in the first orbit of hydrogen?
For hydrogen (Z = 1, n = 1), v_1 = 2.19 x 10^6 m/s. This equals about c/137, where c is the speed of light.
How does electron velocity change from the first to the second orbit?
Since v is proportional to 1/n, the second orbit speed is half the first: v_2 = 2.19x10^6 / 2 = 1.095 x 10^6 m/s for hydrogen.
Is the electron speed the same in all elements?
No. Speed is proportional to Z. For He+ (Z = 2) the n = 1 speed is 4.38 x 10^6 m/s, twice that of hydrogen, because the stronger nuclear charge requires a faster orbit.
What is the ratio of velocity to speed of light for the ground state electron?
v_1/c = 2.19x10^6 / 3x10^8 is about 1/137. This ratio is the fine-structure constant alpha, so the electron moves at roughly 0.73% of the speed of light.
