Physics · Atoms · NEET
m is the electron mass, v is its speed in the orbit, and r is the orbit radius. The product mvr is the angular momentum L of the electron about the nucleus. Bohr said this L cannot take any value — it must equal n times h/2pi. So the smallest allowed angular momentum (n = 1) is h/2pi, the next (n = 2) is 2h/2pi = h/pi, and so on. Nothing in between is allowed.
Bohr introduced it as a postulate (an assumption) to fix Rutherford's unstable atom. He had no proof at first. Later, de Broglie explained it: the electron behaves like a wave, and a stable orbit must fit a whole number of electron wavelengths around the circle. Setting circumference 2pi r = n(lambda) with lambda = h/mv gives 2pi r = n h/(mv), which rearranges exactly to mvr = nh/2pi. So the whole number n comes from fitting whole waves.
n is the principal quantum number and it labels the orbit: n = 1 is the ground (innermost) orbit, n = 2, 3... are higher orbits. It must be a positive integer (1, 2, 3...) because only whole waves fit around the orbit — a half or fractional wave would cancel itself out and the electron could not survive there. n can never be zero or a fraction.
The correct Bohr formula is mvr = nh/2pi. The reduced form uses h-bar (h with a bar), where h-bar = h/2pi, so it is also written mvr = n(h-bar). If you ever see mvr = nh without the 2pi, it is wrong for the Bohr model. In NEET questions, always divide by 2pi.
Compare the given angular momentum with nh/2pi. Example: if L = 1.5(h/pi), rewrite it over 2pi: h/pi = 2(h/2pi), so 1.5(h/pi) = 3(h/2pi). That matches n = 3. General rule: n = (given L) / (h/2pi) = 2pi L / h. Always convert the given value to the form (something) x (h/2pi) and read off n.
The angular momentum of an electron moving in an orbit of hydrogen atom is 1.5(h/pi). The energy in the same orbit is nearly:
A particle of mass m moves around the origin under a constant force F pulling it towards the origin. Using Bohr's model, the radius of the nth orbit and the speed v depend on n as:
Try the real previous-year questions from this chapter — each with the answer and a full solution.
The electron revolves only in those orbits for which its angular momentum is an integer multiple of h/2pi. In symbols: mvr = nh/2pi, where n = 1, 2, 3... This is why the orbits are called 'stationary' or 'allowed' orbits.
For n = 1 (ground state), L = h/2pi = (6.63 x 10^-34)/(2 x 3.14) ≈ 1.05 x 10^-34 J s. This is the smallest angular momentum an electron can have in the Bohr model; it is also called h-bar.
No. The condition mvr = nh/2pi is the same for hydrogen and for hydrogen-like ions such as He+ and Li2+. The atomic number Z affects the orbit radius, speed and energy, but the angular momentum in the nth orbit is always nh/2pi, independent of Z.
Rutherford's model let the electron orbit at any radius, so it would continuously radiate energy, spiral inward and the atom would collapse in about 10^-8 s. Bohr's quantisation picks out a few stable orbits where the electron does not radiate, keeping the atom stable and explaining the sharp line spectrum of hydrogen.
De Broglie showed the electron is a wave with wavelength lambda = h/mv. A stable orbit must contain a whole number of these wavelengths: 2pi r = n(lambda). Putting lambda = h/mv gives 2pi r = nh/mv, which rearranges to mvr = nh/2pi — exactly Bohr's postulate. So quantisation is a natural result of the electron's wave nature.
