Physics · Atoms · NEET
The electron of charge -e goes around a nucleus of charge +Ze in a circle. The nucleus pulls the electron with the Coulomb force F = (1/4 pi e0) * (Z e squared) / r squared. For circular motion this pull must be the centripetal force m v squared / r. Set them equal: (1/4 pi e0)(Z e squared)/r squared = m v squared / r. Cancel one r: (1/4 pi e0)(Z e squared)/r = m v squared. Solve for r: r = Z e squared / (4 pi e0 * m v squared). For hydrogen Z = 1, so r = e squared / (4 pi e0 * m v squared).
Any object moving in a circle needs a real force pointed toward the centre, called the centripetal force. In an atom the only force available is the electric pull of the nucleus on the electron. So that electric (Coulomb) pull must be the thing playing the centripetal role. It is not two separate forces - it is one force (Coulomb) doing the centripetal job. That is why we write Coulomb force = centripetal force.
No. This is the key point. Rutherford's equation r = e squared / (4 pi e0 * m v squared) allows ANY radius, because the speed v can be anything. There is no rule that fixes v, so there is no fixed radius and no fixed energy. Bohr later added a rule (angular momentum m v r = n h / 2 pi) that only allows special speeds, and only then do you get fixed radii like 0.53 Angstrom. So the radius formula is Rutherford; the quantised values are Bohr.
From r = e squared / (4 pi e0 * m v squared), radius is inversely proportional to v squared. A faster electron sits in a smaller, tighter orbit; a slower electron sits in a larger orbit. This is the same idea as a satellite: closer orbits need higher speed. In a NEET numerical you are often given v and asked for r - just plug into r = k e squared / (m v squared) with k = 9 x 10^9.
The force balance (Coulomb = centripetal) and this radius expression belong to Rutherford's classical model - they use only Newton's law and Coulomb's law, no quantisation. Bohr keeps this same force balance but ADDS the quantisation of angular momentum, which then turns v and r into fixed n-dependent values (r proportional to n squared). So the raw formula is Rutherford; the n squared version (r = 0.53 n squared / Z Angstrom) is Bohr.
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. The value of x is: [m_e = 9 x 10^-31 kg, e = 1.6 x 10^-19 C, 1/(4 pi e0) = 9 x 10^9 N m^2 C^-2]
The radius of innermost orbit of hydrogen atom is 5.3 x 10^-11 m. What is the radius of the third allowed orbit of hydrogen atom?
Try the real previous-year questions from this chapter — each with the answer and a full solution.
r = Z e squared / (4 pi e0 * m v squared), where Z is the nuclear charge number (Z = 1 for hydrogen), e is the electron charge, m is its mass and v is its speed. Using 1/(4 pi e0) = k = 9 x 10^9, this is r = k Z e squared / (m v squared).
Because nothing fixes the electron's speed v. Any v gives a valid orbit, so any radius is allowed. Only Bohr's extra rule (quantised angular momentum) picks out special speeds and turns this into fixed radii.
The electrostatic Coulomb attraction between the positive nucleus and the negative electron. That single pull, directed toward the nucleus, plays the role of the centripetal force.
Since r is inversely proportional to v squared, doubling the speed makes the radius one-fourth of its original value. Faster electron means a much tighter orbit.
Yes. Bohr keeps the exact same Coulomb = centripetal balance from Rutherford, then adds m v r = n h / 2 pi. Combining the two gives the famous r_n = 0.53 n squared / Z Angstrom and E_n = -13.6 Z squared / n squared eV.
