Q.The radius of the innermost electron orbit of a hydrogen atom is $5.3 \times 10^{-11}\ \text{m}$. What are the radii of the $n = 2$ and $n = 3$ orbits?

The key idea is that in the Bohr model, the radius of the $n$-th orbit is quantized as $r_n = n^2 r_1$, where $r_1$ is the radius of the innermost orbit ($n=1$).

Given $r_1 = 5.3 \times 10^{-11}\ \text{m}$:

• For $n = 2$: $r_2 = 2^2 \times r_1 = 4 \times 5.3 \times 10^{-11} = 2.12 \times 10^{-10}\ \text{m}$. …

The Bohr model of the hydrogen atom is built on a simple but powerful idea: the electron can only occupy certain allowed orbits where its angular momentum is an integer multiple of $\frac{h}{2\pi}$. This quantization condition leads directly to a neat scaling law for orbit radii.

For a hydrogen-like atom (one electron orbiting a nucleus of charge $+Ze$), the centripetal force comes from the Coulomb attraction:

Combine this with Bohr’s quantization of angular momentum:

Solving these two equations eliminates $v$ and gives the radius of the $n$th orbit:

where $a_0 = 5.3 \times 10^{-11}\ \text{m}$ is the Bohr radius — the radius of the innermost orbit ($n=1$) for hydrogen ($Z=1$).

This is the key: the radius grows as the square of the principal quantum number $n$. No need to re-derive constants each time — once you know $r_1$, all higher orbits follow immediately.

Now let’s apply it step by step.

1. Identify the given data.

   The innermost orbit radius $r_1 = 5.3 \times 10^{-11}\ \text{m}$. This is the Bohr radius for hydrogen.

2. Write the general relation.

   For hydrogen ($Z=1$):

3. Find $r_2$ (for $n=2$). …

Method: Direct Application of Bohr's Radius Formula

Bohr's model tells us that the radius of an electron orbit in a hydrogen atom is proportional to $n^2$, where $n$ is the principal quantum number. The innermost orbit ($n=1$) has a known radius $r_1 = 5.3 \times 10^{-11}\ \text{m}$.

The formula is:

Steps:

1. Identify the given data

   $r_1 = 5.3 \times 10^{-11}\ \text{m}$

   We need $r_2$ and $r_3$.

2. Apply the formula for $n=2$

3. Apply the formula for $n=3$ $r_3 = 3^2 \cdot r_1 = 9 \times (5.3 \times 10^{-11}) = 4.77 \times 10^{-10}\ \text{m}$ …

The most common mistake here is forgetting that the Bohr radius scales as $n^2$, not $n$. Students often multiply the given radius by $n$ instead of $n^2$, so for $n=2$ they write $2 \times 5.3 \times 10^{-11}$ m, and for $n=3$ they write $3 \times 5.3 \times 10^{-11}$ m. That is wrong.

Why it happens: The formula $r_n = n^2 r_1$ comes directly from the quantization condition $m v r = n \hbar$ combined with the Coulomb force balance. The $n^2$ dependence is a key result — it is not obvious from a quick glance, so students misremember it as linear.

How to avoid: Always write the proportionality explicitly before plugging numbers:

Then, for $n=2$:

For $n=3$:

$r_3 = 3^2 \times r_1 = 9 \times 5.3 \times 10^{-11}\ \text{m} = 4.77 \times 10^{-10}\ \text{m}$ …