The ratio of the first excitation energy of a hydrogen atom ($E_2 - E_1$) to its ionisation energy from the ground state is

1. Use $E_n = -13.6/n^2\,\text{eV}$. 2. Ionisation energy from the ground state: $E_\infty - E_1 = 0 - (-13.6) = 13.6\,\text{eV}$. 3. First excitation energy: $E_2 - E_1 = -3.40 - (-13.6) = 10.20\,\text{eV}$. 4. Ratio: $\dfrac{10.20}{13.60} = \dfrac{10.2}{13.6}$. Multiply numerator and denominator by 10: $\dfrac{102}{136} = \dfrac{3}{4}$. 5. So $(E_2 - E_1) : E_\text{ion} = 3 : 4$. 6. Algebraic shortcut: ratio = $\dfrac{(1/1^2) - (1/2^2)}{(1/1^2) - 0} = 1 - \dfrac{1}{4} = \dfrac{3}{4}$. 7. Other options come from computing the wrong difference or comparing $E_2$ to $E_1$ directly. _Source: NCERT Class 12 Physics Part 2, Ch 12, §12.4.1, p. 10 (worked values $E_1 = -13.6$, $E_2 = -3.4$, excitation energy $= 10.2\,\text{eV}$)._