Q.It is found experimentally that $13.6\ \text{eV}$ energy is required to separate a hydrogen atom into a proton and an electron. Compute the orbital radius and the velocity of the electron in a hydrogen atom.

The key idea is that the $13.6\ \text{eV}$ binding energy equals the magnitude of the total energy $E$ of the electron in the ground state of hydrogen. For a circular orbit, the total energy is half the potential energy: $E = -\frac{ke^2}{2r}$.

Step 1: Convert the binding energy to joules.

Since $E = -13.6\ \text{eV}$, we have $|E| = \frac{ke^2}{2r}$.

Step 2: Solve for the orbital radius $r$.

Step 3: Find the velocity using $mv^2/r = ke^2/r^2$, so $v = \sqrt{\frac{ke^2}{mr}}$. …

The 13.6 eV given is not just any number — it is the ionization energy, the minimum energy needed to completely remove the electron from the proton's influence. In the Bohr model, this equals the negative of the total energy of the electron in the $n=1$ orbit. So the electron's total energy in the ground state is $-13.6\ \text{eV}$.

Why does this help? Because in the Bohr model, the total energy, orbital radius, and velocity are all linked by simple relations. If we know one, we can find the others. The trick is to work in SI units consistently — convert eV to joules, then use Coulomb's law and Newton's second law alongside Bohr's angular momentum quantization.

Let's go step by step.

1. Convert the given energy to joules

The ionization energy is $13.6\ \text{eV}$. Since $1\ \text{eV} = 1.602 \times 10^{-19}\ \text{J}$:

This is the magnitude of the total energy $|E|$ of the electron in the ground state. In the Bohr model, the total energy is negative (bound state), so:

2. Recall the Bohr model relations for hydrogen

For an electron in a circular orbit around a proton, two equations hold:

• Newton's second law (Coulomb force provides centripetal acceleration):

• Bohr's quantization of angular momentum (for the ground state, $n=1$):

Here $m = 9.109 \times 10^{-31}\ \text{kg}$ (electron mass), $e = 1.602 \times 10^{-19}\ \text{C}$, $\varepsilon_0 = 8.854 \times 10^{-12}\ \text{F/m}$, and $h = 6.626 \times 10^{-34}\ \text{J·s}$.

3. Use the total energy to find the radius

The total energy is the sum of kinetic and potential energies:

From the centripetal equation, we have $\frac{m v^2}{r} = \frac{1}{4\pi\varepsilon_0} \frac{e^2}{r^2}$, which gives:

So the kinetic energy $K = \frac{1}{2} m v^2 = \frac{1}{2} \cdot \frac{1}{4\pi\varepsilon_0} \frac{e^2}{r}$.

The potential energy $U = -\frac{1}{4\pi\varepsilon_0} \frac{e^2}{r}$.

Therefore:

This is a neat result: the total energy is exactly half the potential energy (and the negative of the kinetic energy). Rearranging for $r$:

Since $E$ is negative, $r$ comes out positive. Plug in the numbers:

First compute $\frac{1}{4\pi\varepsilon_0} = 8.988 \times 10^9\ \text{N·m}^2/\text{C}^2$. Then:

$r = \frac{1}{2} \times 8.988 \times 10^9 \times \frac{2.566 \times 10^{-38}}{2.179 \times 10^{-18}}$ …

Method: Bohr Model for Hydrogen Atom

This problem uses the Bohr model of the hydrogen atom, which combines classical circular motion with quantised angular momentum. The key idea is that the electron orbits the proton in a circle, held by electrostatic attraction, but only certain orbits are allowed — those where the angular momentum is an integer multiple of $\frac{h}{2\pi}$.

Step 1: Write the energy condition

The total energy of the electron in the $n=1$ orbit (ground state) is given as $13.6\ \text{eV}$. This is the ionisation energy — the energy needed to remove the electron completely. In the Bohr model, the total energy is:

where $k = \frac{1}{4\pi\epsilon_0} = 9 \times 10^9\ \text{N m}^2/\text{C}^2$, $e = 1.6 \times 10^{-19}\ \text{C}$, and $r$ is the orbital radius. The negative sign means the electron is bound.

Since $|E| = 13.6\ \text{eV}$, we have:

Convert $13.6\ \text{eV}$ to joules: $13.6 \times 1.6 \times 10^{-19} = 2.176 \times 10^{-18}\ \text{J}$.

Step 2: Solve for the orbital radius $r$

From the energy equation:

Substitute values:

Step 3: Find the velocity $v$

The centripetal force is provided by the electrostatic attraction:

Cancel one $r$:

So: …

Common Mistakes in Rutherford/Bohr Radius & Velocity Problems

This question is a classic — it tests whether you truly understand the link between energy, radius, and velocity in the Bohr model. Most students jump straight to formulas without connecting the given energy to the right physical quantity.

Mistake 1: Confusing the given 13.6 eV with kinetic energy

The most frequent error: students see 13.6 eV and immediately plug it into $E_k = \frac{1}{2}mv^2$ to find velocity. But 13.6 eV is the total energy (or the ionization energy), not the kinetic energy.

Why this happens: The problem says "energy required to separate" — that's the binding energy, which equals the magnitude of the total energy of the electron in the ground state. In the Bohr model:

For $n=1$, $E_{\text{total}} = -13.6\ \text{eV}$. The kinetic energy is $+13.6\ \text{eV}$, and the potential energy is $-27.2\ \text{eV}$.

How to avoid: Always write down the three energy relations side by side before substituting numbers:

• $E_{\text{total}} = -\frac{ke^2}{2r}$ (or $-13.6/n^2$ eV)
• $E_k = \frac{ke^2}{2r} = +13.6\ \text{eV}$ (for $n=1$)
• $E_p = -\frac{ke^2}{r} = -27.2\ \text{eV}$

Mistake 2: Using the wrong value of $e$ (electronic charge)

Students often use $e = 1.6 \times 10^{-19}\ \text{C}$ but forget to square it when it appears as $e^2$ in formulas. Or they mix up $e$ with the charge on the nucleus ($Ze$), but here $Z=1$.

How to avoid: Write the formula with $e^2$ explicitly, then substitute $e = 1.6 \times 10^{-19}\ \text{C}$ and square it immediately. Keep units consistent — use SI units throughout.

Mistake 3: Forgetting to convert eV to joules

The given energy is in electronvolts, but the Coulomb constant $k = 9 \times 10^9$ and other quantities are in SI units. If you plug 13.6 directly without converting to joules, your radius will be off by a factor of $10^{19}$.

How to avoid: Before starting any calculation, convert all energies to joules. Write the conversion step explicitly.

Mistake 4: Using the wrong formula for orbital radius

Some students try to derive radius from $F = \frac{mv^2}{r}$ alone, forgetting they need a second equation (either energy quantization or angular momentum quantization). Without Bohr's postulate, you cannot get a unique radius.

The correct approach: Use the total energy relation:

Since $|E_{\text{total}}| = 13.6\ \text{eV} = 2.176 \times 10^{-18}\ \text{J}$, we have:

Solve for $r$:

Mistake 5: Computing velocity from the wrong energy

After finding $r$, students sometimes use $E_k = \frac{1}{2}mv^2$ but plug in the total energy value (negative) instead of the kinetic energy magnitude. …