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Energy and radius of first Bohr orbit of $\text{He}^+$ and $\text{Li}^{2+}$ are [Given: $R_H = 2.18 \times 10^{-18}\text{ J}$, $a_0 = 52.9\text{ pm}$]:
A
$E_n(\text{Li}^{2+}) = -19.62 \times 10^{-18}\text{ J}; r_n(\text{Li}^{2+}) = 17.6\text{ pm}; E_n(\text{He}^+) = -8.72 \times 10^{-18}\text{ J}; r_n(\text{He}^+) = 26.4\text{ pm}$
B
$E_n(\text{Li}^{2+}) = -8.72 \times 10^{-18}\text{ J}; r_n(\text{Li}^{2+}) = 26.4\text{ pm}; E_n(\text{He}^+) = -19.62 \times 10^{-18}\text{ J}; r_n(\text{He}^+) = 17.6\text{ pm}$
C
$E_n(\text{Li}^{2+}) = -19.62 \times 10^{-16}\text{ J}; r_n(\text{Li}^{2+}) = 17.6\text{ pm}; E_n(\text{He}^+) = -8.72 \times 10^{-16}\text{ J}; r_n(\text{He}^+) = 26.4\text{ pm}$
D
$E_n(\text{Li}^{2+}) = -8.72 \times 10^{-16}\text{ J}; r_n(\text{Li}^{2+}) = 17.6\text{ pm}; E_n(\text{He}^+) = -19.62 \times 10^{-16}\text{ J}; r_n(\text{He}^+) = 17.6\text{ pm}$
Explanation
$E_n = -R_H \frac{Z^2}{n^2}$ and $r_n = a_0 \frac{n^2}{Z}$. For $\text{He}^+$ ($Z=2$): $E_1 = -8.72 \times 10^{-18}\text{ J}$, $r_1 = 26.45\text{ pm}$. For $\text{Li}^{2+}$ ($Z=3$): $E_1 = -19.62 \times 10^{-18}\text{ J}$, $r_1 = 17.63\text{ pm}$.
Detailed Solution
For hydrogenic species: Energy is $E_n = -R_H \frac{Z^2}{n^2}$ and radius is $r_n = a_0 \frac{n^2}{Z}$. For $\text{He}^+$ ($Z = 2, n = 1$): $E_1 = -2.18 \times 10^{-18} \times (2^2) = -8.72 \times 10^{-18}\text{ J}$, $r_1 = \frac{52.9}{2} = 26.45\text{ pm} \approx 26.4\text{ pm}$. For $\text{Li}^{2+}$ ($Z = 3, n = 1$): $E_1 = -2.18 \times 10^{-18} \times (3^2) = -19.62 \times 10^{-18}\text{ J}$, $r_1 = \frac{52.9}{3} = 17.63\text{ pm} \approx 17.6\text{ pm}$.