UPSC NDA 2 2017 General Ability Solved Paper

Concept:The ionization energy is the minimum energy needed to remove an electron from the ground state of an atom to infinity.Explanation:For a hydrogen atom, the energy of an electron in the $n^{\text{th}}$ orbit is given by the formula:$$E_n = -13.6 \frac{Z^2}{n^2} \ \text{eV}$$Here, $Z = 1$ for hydrogen and the ground state corresponds to $n = 1$.Substituting these values gives the ground state energy:$$E_1 = -13.6 \frac{1^2}{1^2} = -13.6 \ \text{eV}$$When the electron is completely removed from the atom, its energy is taken as zero at infinity:$$E_\infty = 0$$The ionization energy is the difference between the final energy (at infinity) and the initial energy (ground state):$$\text{Ionization energy} = E_\infty - E_1 = 0 - (-13.6) = 13.6 \ \text{eV}$$Thus, the correct value is $13.6 \ \text{eV}$, which matches option B.The other options are incorrect: $13.6 \ \text{MeV}$ is too large, $13.6 \ \text{Joules}$ is far too large, and zero is impossible because energy is required to overcome attraction.Answer:B. 13.6 eV