Q.(a) What are postulates of Bohr's model of Hydrogen atom?

The Problem Bohr Solved

Before Bohr, the atom was a puzzle. Rutherford had shown that atoms have a tiny, dense nucleus with electrons orbiting around it — like a miniature solar system. But there was a fatal flaw in that picture.

Classical physics said an accelerating charged particle must radiate energy. An electron orbiting a nucleus is constantly accelerating (changing direction), so it should continuously lose energy, spiral into the nucleus, and collapse. That doesn't happen. Atoms are stable. Also, atoms emit light only at specific, discrete wavelengths — not a continuous rainbow. The solar-system model couldn't explain either fact.

Bohr needed a new idea. He kept the planetary picture but added rules that broke from classical physics — rules that matched what experiments actually showed.

The Core Intuition

Think of an electron like a marble rolling on a staircase, not a ramp. On a ramp, the marble can stop at any height. On a staircase, it can only rest on the steps — never in between. Bohr said electrons can only occupy certain allowed "steps" (orbits) around the nucleus. While on a step, the electron does not radiate energy. It only emits or absorbs light when it jumps from one step to another, and the light's colour (wavelength) corresponds exactly to the energy difference between the steps.

This is why atomic spectra show sharp lines: each line is a jump between two specific steps.

The Precise Postulates

Bohr's model rests on three postulates, stated clearly:

What the Model Predicts

From these postulates, Bohr derived explicit formulas. For a hydrogen-like atom (one electron, nuclear charge $Ze$):

Radius of the $n$th orbit:

where $a_0 = 0.529 \, \text{Å}$ is the Bohr radius (the smallest orbit, for $n=1$ in hydrogen).

Energy of the $n$th orbit:

The negative sign means the electron is bound to the nucleus. The ground state ($n=1$) has energy $-13.6 \, \text{eV}$ for hydrogen.

Wavelength of emitted light when an electron jumps from $n_i$ to $n_f$ ($n_i > n_f$):

where $R = \frac{m e^4}{8 \varepsilon_0^2 h^3 c} \approx 1.097 \times 10^7 \, \text{m}^{-1}$ is the Rydberg constant. …