Q.The ground state energy of hydrogen atom is $-13.6\ \text{eV}$. What are the kinetic and potential energies of the electron in this state?

Concept: Bohr Model Energy Levels

In the Bohr model, the total energy $E = K + U$ for a hydrogen atom is $E = -13.6\ \text{eV}$ in the ground state. The virial theorem for a Coulomb potential gives $K = -E$ and $U = 2E$.

Reasoning

1. For an electron in a stable circular orbit around a proton, the centripetal force is provided by the electrostatic attraction: $\frac{mv^2}{r} = \frac{ke^2}{r^2}$.
2. Kinetic energy $K = \frac{1}{2}mv^2 = \frac{ke^2}{2r}$. Potential energy $U = -\frac{ke^2}{r}$. …

The Bohr model of the hydrogen atom gives us a beautifully simple relationship between the electron's total, kinetic, and potential energies. The key insight is that the electron is held in a circular orbit by the electrostatic attraction to the proton, and the centripetal force condition leads to a direct proportionality between these energies.

For any stable circular orbit in an inverse-square force law (like Coulomb's law), the kinetic energy is exactly half the magnitude of the potential energy, but with opposite sign. This is a special case of the virial theorem for a $1/r$ potential.

Let's verify this from first principles.

1. Write the total energy. The electron has kinetic energy $K = \frac{1}{2}mv^2$ and potential energy $U = -\frac{ke^2}{r}$ (taking $U=0$ at infinity). So:

2. Apply the centripetal force condition. For a stable circular orbit, the electrostatic force provides the necessary centripetal acceleration:

Multiply both sides by $r$:

3. Relate kinetic energy to potential. Notice that $mv^2 = 2K$. So:

But $\frac{ke^2}{r} = -U$ (since $U = -\frac{ke^2}{r}$). Therefore:

4. Express $K$ and $U$ in terms of $E$. From $E = K + U$ and $U = -2K$, substitute: $E = K - 2K = -K \quad \Rightarrow \quad K = -E$ …

Method: Using the Bohr model relations for a hydrogen-like atom.

In the Bohr model, the electron moves in a circular orbit around the nucleus. The total energy $E$ is the sum of kinetic energy $K$ and potential energy $U$:

For a Coulomb force, the potential energy is $U = -\frac{ke^2}{r}$ and the kinetic energy is $K = \frac{1}{2}mv^2$. From the centripetal force condition $\frac{mv^2}{r} = \frac{ke^2}{r^2}$, we get:

Comparing with $U = -\frac{ke^2}{r}$, we see:

So the total energy becomes:

This gives $U = 2E$ and $K = -E$.

Step-by-step for the given problem:

1. The ground state energy is given: $E = -13.6\ \text{eV}$.

2. Using $K = -E$:

   $K = -(-13.6\ \text{eV}) = +13.6\ \text{eV}$.

3. Using $U = 2E$: …

Students often know the formula $E_n = -\frac{13.6}{n^2}\ \text{eV}$ but then fumble when asked for kinetic or potential energy separately. Here are the most common mistakes and how to avoid each.

Mistake 1: Confusing total energy with kinetic energy

Many students see $-13.6\ \text{eV}$ and think that's the kinetic energy, or they assume kinetic energy is also negative. In the Bohr model, kinetic energy is always positive — the electron is moving, so $K > 0$.

How to avoid: Remember the key relationship derived from the Bohr model:

• Total energy: $E = K + U$
• For a circular orbit, the virial theorem for a Coulomb force gives $K = -E$ and $U = 2E$.

Since $E = -13.6\ \text{eV}$, you get:

Mistake 2: Forgetting that potential energy is negative and twice the total energy

Students sometimes write $U = E - K$ and then plug $K = 13.6\ \text{eV}$, getting $U = -13.6 - 13.6 = -27.2\ \text{eV}$ — which is correct, but they don't understand why it's double. Without the reasoning, they might misapply it in other contexts.

How to avoid: Derive it once from force balance:

• Centripetal force: $\frac{mv^2}{r} = \frac{ke^2}{r^2}$
• Kinetic energy: $K = \frac12 mv^2 = \frac{ke^2}{2r}$
• Potential energy: $U = -\frac{ke^2}{r}$
• So $U = -2K$, and $E = K + U = K - 2K = -K$.

From $E = -K$, you get $K = -E$ and $U = 2E$. This is a clean, exam-ready chain.

Mistake 3: Using $n=1$ but forgetting the sign convention

Some students write $E_1 = -\frac{13.6}{1^2} = -13.6\ \text{eV}$, then say $K = \frac{13.6}{1^2} = 13.6\ \text{eV}$ — which works here, but they don't generalise. For $n=2$, they might incorrectly write $K = \frac{13.6}{2^2} = 3.4\ \text{eV}$, when actually $K = -E = 3.4\ \text{eV}$ (same number, but the reasoning is fragile).

How to avoid: Always use the relationships $K = -E$ and $U = 2E$, not a separate formula for each $n$. That way, for any state:

$U_n = 2E_n = -\frac{27.2}{n^2}\ \text{eV}$ …