Question 1

According to Bohr's model, the kinetic energy ($KE$) of an electron in an orbit is given by: $KE = -E_{total} = 13.6 \times \frac{Z^2}{n^2} \text{ eV}$ where $Z$ is the atomic number and $n$ is the orbit number. Calculating for each case: (A) For first orbit of H atom: $Z = 1, n = 1 \Rightarrow KE \propto \frac{1^2}{1^2} = 1$ (B) For first orbit of $He^+$: $Z = 2, n = 1 \Rightarrow KE \propto \frac{2^2}{1^2} = 4$ (C) For second orbit of $He^+$: $Z = 2, n = 2 \Rightarrow KE \propto \frac{2^2}{2^2} = 1$ (D) For second orbit of $Li^{2+}$: $Z = 3, n = 2 \Rightarrow KE \propto \frac{3^2}{2^2} = 2.25$ The highest value is 4, which corresponds to the first orbit of $He^+$. Hence, the correct option is (B).