Physics · Dual Nature Of Radiation And Matter · NEET
The electron in an orbit is a standing wave, not a moving ball. For a wave wrapped around a closed circle to be stable, its crest and trough must line up when it comes back to the start point. This only happens if the length of the orbit (the circumference 2*pi*r) is an exact whole-number multiple of the wavelength. So 2*pi*r_n = n*lambda, where n = 1, 2, 3... If it were not a whole number, the wave would overlap itself out of step and cancel (destructive interference), so no stable orbit forms.
Start from 2*pi*r = n*lambda and put lambda = h/p = h/(mv). Then 2*pi*r = n*h/(mv). Rearranging gives mvr = n*h/(2*pi). That is exactly Bohr's quantisation of angular momentum, L = n*h/(2*pi). So de Broglie's wave idea gives a physical reason WHY angular momentum is quantised - Bohr just assumed it, de Broglie explained it.
Exactly n. In the first orbit (n = 1) one full wavelength fits around. In the second orbit (n = 2) two full wavelengths fit, and so on. So the number of wavelengths in an orbit equals its principal quantum number n. This is a very common NEET one-liner.
Yes. lambda_n = 2*pi*r_n / n. For hydrogen r_n = n^2 * a_0, so lambda_n = 2*pi*(n^2*a_0)/n = 2*pi*n*a_0. This means lambda is directly proportional to n - the wavelength gets larger in higher orbits (lambda_2 = 2*lambda_1, lambda_3 = 3*lambda_1). Do not assume it stays constant.
Use lambda_n = 2*pi*n*a_0 with a_0 = 0.053 nm. For n = 1: lambda = 2*pi*(1)(0.053) = 0.33 nm. For n = 2: lambda = 2*pi*(2)(0.053) = 0.67 nm. For n = 3: lambda = 2*pi*(3)(0.053) = 1.0 nm. This one shortcut answers the NEET 2025 question directly.
The de Broglie wavelength of an electron in the n = 2 state of the hydrogen atom is close to (Given Bohr radius a_0 = 0.053 nm):
Try the real previous-year questions from this chapter — each with the answer and a full solution.
It is the wavelength of the electron treated as a wave in its orbit. For the nth orbit lambda_n = 2*pi*r_n / n, and for hydrogen this simplifies to lambda_n = 2*pi*n*a_0.
A whole number of de Broglie wavelengths must fit exactly around the orbit: 2*pi*r_n = n*lambda, with n = 1, 2, 3... This makes the electron wave a stable standing wave.
Exactly one full de Broglie wavelength fits around the first orbit. In general n wavelengths fit in the nth orbit.
Larger. Since lambda_n = 2*pi*n*a_0, the wavelength grows in direct proportion to n, so the n = 3 wavelength is three times the n = 1 wavelength.
Putting lambda = h/(mv) into 2*pi*r = n*lambda gives mvr = n*h/(2*pi). This is Bohr's rule L = n*h/(2*pi), so the wave picture provides the reason behind it.
