Match List - I with List – II.
| List - I | List – II |
| (Total energy eV) | (States of electron in Hydrogen atom) |
| 1. \( -3.40 \) | (I) At infinity |
| 2. \( -13.6 \) | (II) I excited state |
| 3. 0 | (III) Ground state |
| 4. \( -0.85 \) | (IV) III excited state |
Choose the correct answer from the options given below :
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (1) - (II), (2) – (III), (3) – (I), (4) – (IV)
In the Bohr model the total energy of the electron in the \( n \)th orbit of hydrogen is
\( \displaystyle E_n = -\frac{13.6}{n^{2}}\,\mathrm{eV} \)
The ground state is \( n = 1 \), the first excited state is \( n = 2 \), the third excited state is \( n = 4 \), and at infinity (\( n \rightarrow \infty \)) the energy is zero.
(1) \( -3.40 \) eV — this matches (II) I excited state. \( E_2 = -13.6/4 = -3.40\,\mathrm{eV} \).
(2) \( -13.6 \) eV — this matches (III) Ground state. \( E_1 = -13.6/1 = -13.6\,\mathrm{eV} \), the lowest energy.
(3) 0 — this matches (I) At infinity. A free electron at rest far from the nucleus has zero total energy; this is the ionisation limit.
(4) \( -0.85 \) eV — this matches (IV) III excited state. \( E_4 = -13.6/16 = -0.85\,\mathrm{eV} \).
NCERT (2026-27):
Bohr Model of the Hydrogen Atom – “… Since 1 eV = 1.6 × 10–19 J, Eq. (12.9) can be rewritten as
\( \displaystyle E_n = -\frac{13.6\ \mathrm{eV}}{n^2} \qquad (12.10) \)”
Energy Levels – “… The lowest state of the atom, called the ground state, is that of the lowest energy, with the electron revolving in the orbit of smallest radius, the Bohr radius, a0. The energy of this state (n = 1), E1 is –13.6 eV. … From Eq. (12.10), for n = 2; the energy E2 is –3.40 eV. It means that the energy required to excite an electron in hydrogen atom to its first excited state, is an energy equal to E2 – E1 = –3.40 eV – (–13.6) eV = 10.2 eV.”