Calculate radius of 2nd orbit of hydrogen atom using Bohr model

Question

Using Bohr’s model of the atom, calculate the radius of the second orbit of the hydrogen atom.

Solution — Step by Step

According to Bohr’s model, the radius of the $n$ -th orbit of a hydrogen-like atom is given by:

r_n = \frac{n^2 h^2 \varepsilon_0}{\pi m e^2 Z}

For hydrogen ( $Z = 1$ ), this simplifies to:

r_n = \frac{n^2 h^2 \varepsilon_0}{\pi m e^2}

The standard form commonly used in calculations:

r_n = n^2 \times r_1

where $r_1 = a_0 = 0.529\ \text{Å} = 0.529 \times 10^{-10}\ \text{m}$ is the Bohr radius (radius of the first orbit).

For the second orbit, $n = 2$ :

r_2 = n^2 \times a_0 = (2)^2 \times 0.529\ \text{Å}

r_2 = 4 \times 0.529\ \text{Å}

r_2 = 2.116\ \text{Å} \approx 2.12\ \text{Å}

In SI units:

r_2 = 4 \times 0.529 \times 10^{-10}\ \text{m} = 2.116 \times 10^{-10}\ \text{m}

Radius of the 2nd orbit = 2.116 Å

Bohr’s key postulate: the angular momentum of the electron is quantised in integral multiples of $\dfrac{h}{2\pi}$ :

m v r = \frac{n h}{2\pi}

Also, the centripetal force equals the electrostatic force:

\frac{mv^2}{r} = \frac{e^2}{4\pi\varepsilon_0 r^2}

From these two equations:

v = \frac{e^2}{2\varepsilon_0 n h} \quad \text{(quantised speeds)}

Substituting back into the angular momentum equation:

r_n = \frac{n^2 h^2 \varepsilon_0}{\pi m e^2} = n^2 \times 0.529\ \text{Å}

For a hydrogen-like ion with atomic number Z (He⁺, Li²⁺, etc.):

r_n = \frac{n^2}{Z} \times 0.529\ \text{Å}

So for He⁺ ( $Z = 2$ ) in the second orbit: $r_2 = \dfrac{4}{2} \times 0.529 = 2 \times 0.529 = 1.058\ \text{Å}$ — smaller than hydrogen’s first orbit!

Higher nuclear charge pulls the electron closer to the nucleus.

Why This Works

The $n^2$ dependence arises because increasing the principal quantum number ( $n$ ) increases both the angular momentum and the orbital size. The electron in the second orbit moves at a lower speed (half the first orbit speed) but farther from the nucleus.

The Bohr radius ( $a_0 = 0.529$ Å) is a fundamental constant of atomic physics — it sets the scale of atomic size. Any atom’s orbital radii are integer-squared multiples of this scale (for hydrogen).

Alternative Method

We can also use the velocity formula to cross-check. The speed of an electron in the $n$ -th orbit is $v_n = \dfrac{v_1}{n}$ where $v_1 = 2.19 \times 10^6\ \text{m/s}$ . Then use $r_n = \dfrac{mv_n^2}{\text{centripetal force}}$ to re-derive — but the direct formula $r_n = n^2 \times 0.529$ Å is much faster.

In JEE and NEET, four key numbers from Bohr’s model are asked repeatedly: $r_1 = 0.529$ Å, $v_1 = 2.19 \times 10^6$ m/s, $E_1 = -13.6$ eV (ground state energy of H), and the frequency of emitted radiation using $\Delta E = h\nu$ . Know all four cold.

Common Mistake

Students often confuse the formula $r_n = n^2 a_0$ (for hydrogen, $Z = 1$ ) with the general formula $r_n = \dfrac{n^2}{Z} a_0$ (for hydrogen-like atoms). For a question about hydrogen, both give the same answer since $Z = 1$ . But for He⁺ or Li²⁺, forgetting the $Z$ in the denominator is a full-mark error.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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