MHTCET 15 May 2024 Shift 1 Paper

Concept:For a hydrogen-like species, the radius of the $n$-th orbit is directly proportional to $n^2$ and inversely proportional to the atomic number $Z$.Explanation:The radius of the $n$-th orbit of a hydrogen-like atom is given by:$$r_n = \frac{n^2 \cdot a_0}{Z}$$Here, $a_0 = 52.9\ \mathrm{pm}$ is the Bohr radius for hydrogen ($n=1, Z=1$).For $\mathrm{Li}^{++}$, the atomic number is $Z = 3$.For the first orbit, $n = 1$.Substitute the values:$$r_1 = \frac{1^2 \cdot 52.9\ \mathrm{pm}}{3}$$$$r_1 = \frac{52.9}{3}\ \mathrm{pm} = 17.63\ \mathrm{pm}$$Answer:The radius of the first orbit of $\mathrm{Li}^{++}$ is $17.63\ \mathrm{pm}$, which matches Option C.