State the postulates of Bohr's model of the hydrogen atom. Using these postulates, derive the expression for the radius of the $n$-th orbit of a hydrogen atom and show that the radius is directly proportional to the square of the principal quantum number ($n^2$).

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Postulates of Bohr's Model of Hydrogen Atom 
Bohr's model of the hydrogen atom is based on the following three fundamental postulates: Nuclear Concept: An atom consists of a central positive core called the nucleus, where the entire positive charge and almost the entire mass of the atom are concentrated. The electrons revolve around the nucleus in circular paths called orbits, the necessary centripetal force being provided by the electrostatic force of attraction between the positively charged nucleus and the negatively charged electrons. Quantization of Angular Momentum: Electrons can only revolve in certain stable, non-radiating orbits called stationary orbits. In these orbits, the orbital angular momentum ($L$) of the electron is an integral multiple of $\frac{h}{2\pi}$, where $h$ is Planck's constant. Mathematically, $L = mvr = \frac{nh}{2\pi}$, where $n = 1, 2, 3, \dots$ (principal quantum number), $m$ is the mass of the electron, $v$ is its velocity, and $r$ is the radius of the orbit. Frequency Condition: An electron does not radiate energy while revolving in a stationary orbit. Energy is radiated or absorbed only when an electron jumps from one allowed orbit to another. If $E1$ and $E2$ are the energies associated with these orbits, the frequency $
u$ of the emitted or absorbed radiation is given by $h
u = E2 - E1$. Derivation for the Radius of the $n$-th Orbit 
Let us consider an electron of mass $m$ and charge $e$ revolving around a nucleus of atomic number $Z$ (for hydrogen, $Z = 1$) with velocity $v$ in an orbit of radius $r$. Step 1: Balancing Forces The electrostatic force of attraction between the nucleus and the electron provides the necessary centripetal force: $$\frac{1}{4\pi\varepsilon0} \cdot \frac{(Ze)(e)}{r^2} = \frac{mv^2}{r}$$ $$\frac{1}{4\pi\varepsilon0} \cdot \frac{Ze^2}{r^2} = \frac{mv^2}{r}$$ $$\frac{1}{4\pi\varepsilon0} \cdot \frac{Ze^2}{r} = mv^2 \quad 	ext{--- (Equation 1)}$$ Step 2: Applying Bohr's Quantization Condition According to Bohr's second postulate: $$mvr = \frac{nh}{2\pi}$$ Solving for velocity $v$: $$v = \frac{nh}{2\pi mr} \quad 	ext{--- (Equation 2)}$$ Step 3: Substituting Velocity into Equation 1 Substitute the value of $v$ from Equation 2 into Equation 1: $$\frac{1}{4\pi\varepsilon0} \cdot \frac{Ze^2}{r} = m \left( \frac{nh}{2\pi mr} ight)^2$$ $$\frac{1}{4\pi\varepsilon0} \cdot \frac{Ze^2}{r} = m \cdot \frac{n^2 h^2}{4\pi^2 m^2 r^2}$$ $$\frac{1}{4\pi\varepsilon0} \cdot \frac{Ze^2}{r} = \frac{n^2 h^2}{4\pi^2 m r^2}$$ Step 4: Solving for Radius ($r$) Rearranging the terms to solve for $r$: $$r = \frac{n^2 h^2 \cdot 4\pi\varepsilon0}{4\pi^2 m Z e^2}$$ $$rn = \frac{n^2 h^2 \varepsilon0}{\pi m Z e^2}$$ Conclusion 
For a hydrogen atom, $Z = 1$. Substituting the values of constant terms ($h$, $\varepsilon0$, $\pi$, $m$, $e$), we get the radius of the first orbit ($r1 \approx 0.53 	ext{ \AA}$). Since $h$, $\varepsilon0$, $\pi$, $m$, and $e$ are all fundamental constants, we can clearly see that: $$rn \propto n^2$$
Thus, the radius of the $n$-th orbit is directly proportional to the square of the principal quantum number $n$.