Physics · Atoms · NEET
The electron moves in a circle of radius r with speed v. One full round covers the circumference 2 pi r, so the time period is T = distance / speed = 2 pi r / v. This is just uniform circular motion. Put the Bohr values r = 0.53 n squared / Z angstrom and v = 2.18 x 10^6 (Z / n) m/s to get T for any orbit.
Frequency of revolution f is the number of complete rounds the electron makes in one second. It is simply f = 1 / T = v / (2 pi r). Do not confuse this f (how many times per second the electron circles the nucleus) with the frequency of a photon emitted during a jump between energy levels. They are different quantities.
For hydrogen, r is proportional to n squared and v is proportional to 1/n. So T = 2 pi r / v is proportional to n squared / (1/n) = n cubed. Therefore T is proportional to n cubed, and f = 1/T is proportional to 1 / n cubed. Higher orbits mean a much longer time period. Going from n = 1 to n = 2 makes T eight times larger (2 cubed = 8).
Include Z for hydrogen-like ions. Since r is proportional to n squared / Z and v is proportional to Z / n, T is proportional to n cubed / Z squared and f is proportional to Z squared / n cubed. So for the same n, a larger nuclear charge Z makes the electron revolve much faster (f grows as Z squared).
For n = 1, Z = 1: v = 2.18 x 10^6 m/s and r = 0.53 x 10^-10 m. Then f = v / (2 pi r) = (2.18 x 10^6) / (2 x 3.14 x 0.53 x 10^-10) which is about 6.6 x 10^15 revolutions per second (Hz). This matches the NCERT worked example. The time period T = 1/f is about 1.5 x 10^-16 s.
Only in classical physics were they assumed equal. In the real Bohr atom, an electron in a stable orbit revolves without radiating, so its revolution frequency is NOT the frequency of emitted light. Light is emitted only when the electron jumps between orbits, and that photon frequency comes from f_photon = (E_i minus E_f) / h, a completely separate formula.
Try the real previous-year questions from this chapter — each with the answer and a full solution.
T = 2 pi r / v. Using Bohr values it becomes T proportional to n cubed / Z squared. For hydrogen ground state T is about 1.5 x 10^-16 seconds.
About 6.6 x 10^15 Hz (revolutions per second), found from f = v / (2 pi r) with v = 2.18 x 10^6 m/s and r = 0.53 x 10^-10 m.
Because T = 2 pi r / v, and r is proportional to n squared while v is proportional to 1/n. Dividing n squared by 1/n gives n cubed.
f = 1/T, and since T is proportional to n cubed, f is proportional to 1 / n cubed. The electron in higher orbits circles the nucleus far fewer times per second.
No. A Bohr electron in a stable orbit does not radiate. Photon frequency comes only from energy-level jumps using f = (E_i minus E_f) / h.
