Physics · Atoms · NEET
Start from the standing-wave condition: the circumference must hold a whole number of waves, so 2*pi*r_n = n*lambda_n. Rearranging gives lambda_n = 2*pi*r_n / n. Put in r_n = a0*n^2 (a0 = 0.529 angstrom = 52.9 pm) and it becomes lambda_n = 2*pi*a0*n. So the wavelength is directly proportional to n.
De Broglie treated the orbiting electron as a wave on a closed loop. For the wave to survive (not cancel itself), it must join smoothly end to end. That resonance condition allows only whole numbers of wavelengths around the loop: 2*pi*r = n*lambda. The first orbit fits exactly 1 wave, the second fits 2 waves, and so on. This is the physical reason behind Bohr's quantisation of angular momentum.
It increases. Because lambda_n = 2*pi*a0*n, doubling n doubles the wavelength. Students often confuse this with speed, which decreases (v_n proportional to 1/n), or with energy, which becomes less negative. Only the wavelength grows in a simple straight line with n.
They are not the same quantity. Radius r_n = a0*n^2 grows as n squared. Wavelength lambda_n = 2*pi*a0*n grows only as n. Their ratio is 2*pi*r_n / lambda_n = n, which just tells you the number of waves. So a larger orbit is not the same shape scaled up; the number of waves in it also grows.
Exactly n waves. In the first orbit (n=1) there is 1 complete wave, in the second (n=2) there are 2, in the fifth (n=5) there are 5. This is just the standing-wave condition 2*pi*r_n = n*lambda_n read directly. It is a favourite one-line NEET answer.
Yes. lambda = h / p = h / (m*v_n). Using v_n = 2.19e6 / n m/s, you can compute lambda for any n. It gives the same answer as 2*pi*a0*n, but the 2*pi*a0*n route is far faster in an exam because it skips the velocity step.
In a hydrogen atom, the de Broglie wavelength of an electron in the second Bohr orbit is: (Given Bohr radius a0 = 52.9 pm)
The de Broglie wavelength of an electron in the n = 2 state of a hydrogen atom is close to: (Given Bohr radius = 0.052 nm)
Try the real previous-year questions from this chapter — each with the answer and a full solution.
For n = 1, lambda_1 = 2*pi*a0*1 = 2*pi*(0.529 angstrom) = 3.32 angstrom (about 0.33 nm). This is the smallest de Broglie wavelength among all the orbits, since wavelength grows with n.
The plain form 2*pi*a0*n is only for hydrogen (Z = 1). For a hydrogen-like ion the radius is r_n = a0*n^2 / Z, so lambda_n = 2*pi*r_n / n = 2*pi*a0*n / Z. The wavelength shrinks by a factor Z.
Substituting lambda = h/(m*v_n) into 2*pi*r_n = n*lambda gives 2*pi*r_n = n*h/(m*v_n), which rearranges to m*v_n*r_n = n*h/(2*pi). That is exactly Bohr's quantisation of angular momentum. The wave picture is the reason behind that rule.
No, in the form lambda_n = 2*pi*a0*n the mass is already absorbed into a0. If you instead use lambda = h/(m*v_n), the mass appears explicitly but the final number is the same for the electron.
