Bohr's Theory of Hydrogen Atom: Derivation

Step 1 (Bohr's Postulates): [1 Mark]

The three postulates of Bohr's theory are mathematically expressed as:

• Postulate 1 (Coulombic Force): The centripetal force required for circular motion is provided by the electrostatic force of attraction: $$\frac{mv^2}{r} = \frac{1}{4\pi\epsilon_0} \frac{Ze^2}{r^2}$$
• Postulate 2 (Quantization of Angular Momentum): The electron revolves only in orbits where its angular momentum is an integral multiple of $\frac{h}{2\pi}$: $$mvr = \frac{nh}{2\pi}$$
• Postulate 3 (Frequency Condition): The energy emitted or absorbed during a transition between two energy levels is: $$h\nu = E_f - E_i$$

Step 2 (Mathematical Working): [1.5 Marks]

(a) Radius of the orbit ($r \propto n^2$):

From Postulate 2, $v = \frac{nh}{2\pi mr}$. Substituting this into the Postulate 1 equation:

$$\frac{m}{r} \left( \frac{nh}{2\pi mr} \right)^2 = \frac{1}{4\pi\epsilon_0} \frac{Ze^2}{r^2}$$ $$\frac{m}{r} \cdot \frac{n^2h^2}{4\pi^2m^2r^2} = \frac{Ze^2}{4\pi\epsilon_0r^2}$$

Solving for $r$:

$$r = \frac{n^2h^2\epsilon_0}{\pi mZe^2} \implies r \propto n^2$$

(b) Total energy of the atom ($E \propto 1/n^2$):

Total energy $E = K.E. + P.E.$

$$E = \frac{1}{2}mv^2 - \frac{1}{4\pi\epsilon_0} \frac{Ze^2}{r}$$

Using $\frac{mv^2}{r} = \frac{1}{4\pi\epsilon_0} \frac{Ze^2}{r^2}$, we get $K.E. = \frac{1}{2} \left( \frac{1}{4\pi\epsilon_0} \frac{Ze^2}{r} \right)$.

$$E = \frac{1}{8\pi\epsilon_0} \frac{Ze^2}{r} - \frac{1}{4\pi\epsilon_0} \frac{Ze^2}{r} = -\frac{Ze^2}{8\pi\epsilon_0r}$$

Substituting $r \propto n^2$ into the energy expression:

$$E = -\frac{Ze^2}{8\pi\epsilon_0} \left( \frac{\pi mZe^2}{n^2h^2\epsilon_0} \right) = -\frac{mZ^2e^4}{8\epsilon_0^2h^2n^2}$$

Step 3 (Final Concluded Answer): [0.5 Mark]

Since $m, Z, e, \epsilon_0,$ and $h$ are constants for a hydrogen-like atom:

(a) $r_n = \left( \frac{h^2\epsilon