In the Bohr model of the hydrogen atom, the energy of an electron in the $n^\text{th}$ orbit is given by $E_n = -13.6/n^2\,\text{eV}$. The MINIMUM energy required to IONISE a hydrogen atom from its ground state is
A$3.4\,\text{eV}$
B$10.2\,\text{eV}$
C$13.6\,\text{eV}$
D$27.2\,\text{eV}$
Answer & Solution
Correct answer: C. $13.6\,\text{eV}$
1. Ionisation means removing the electron completely, i.e. taking it to $n = \infty$ where $E_\infty = 0$.
2. Ground state is $n = 1$: $E_1 = -13.6/1^2 = -13.6\,\text{eV}$.
3. Minimum energy to ionise from ground state = $E_\infty - E_1 = 0 - (-13.6) = +13.6\,\text{eV}$.
4. Option A is $|E_2|$ — the energy of the first excited state, not the ionisation energy. Option B is the first excitation energy $E_2 - E_1 = 10.2\,\text{eV}$. Option D doubles the ionisation energy.
5. NCERT confirms (§12.4.1): the experimental ionisation energy of hydrogen is in EXCELLENT agreement with Bohr's $13.6\,\text{eV}$ prediction.
_Source: NCERT Class 12 Physics Part 2, Ch 12, §12.4.1 (Energy Levels), p. 10._
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