Question:medium

What is the energy of the ground state of a hydrogen atom according to the Bohr model?

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Energy levels of hydrogen follow \[ E_n = -\frac{13.6}{n^2}\ \text{eV} \] The ground state (\(n=1\)) has energy \(-13.6\ \text{eV}\).
Updated On: Mar 25, 2026
  • \(-13.6\ \text{eV}\)
  • \(-3.4\ \text{eV}\)
  • \(-1.51\ \text{eV}\)
  • \(0\ \text{eV}\)
Show Solution

The Correct Option is A

Solution and Explanation

Step 1: Understanding the Concept:
In the Bohr model of the hydrogen atom, an electron orbits the nucleus in fixed, quantized circular paths called energy levels or shells. The ground state is the lowest energy level closest to the nucleus.
Step 2: Key Formula or Approach:
The energy of an electron in the $n$-th energy level of a hydrogen-like atom is given by the formula:
\[ E_n = -\frac{13.6 Z^2}{n^2} \text{ eV} \]
where $Z$ is the atomic number (for hydrogen, $Z = 1$) and $n$ is the principal quantum number.
Step 3: Detailed Explanation:
The ground state corresponds to the first orbit, so the principal quantum number is $n = 1$.
Substituting $Z = 1$ and $n = 1$ into the energy formula gives:
\[ E_1 = -\frac{13.6 \times (1)^2}{(1)^2} \text{ eV} \]
\[ E_1 = -13.6 \text{ eV} \]
The negative sign signifies that the electron is bound to the nucleus. An external energy of $+13.6 \text{ eV}$ must be supplied to completely remove the electron from this state (ionization energy).
Step 4: Final Answer:
The correct option is (A).
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