Q1 / 160 · AP EAMCET 2019 (21 Apr Shift 1) · Chemistry
The energies of an electron in first orbit of $\mathrm{He}^{+}$and in third orbit of $\mathrm{Li}^{2+}$ in $\mathrm{J}$ are respectively
- Correct answerA$-8.72 \times 10^{-18},-2.18 \times 10^{-18}$
- B$-8.72 \times 10^{-18},-1.96 \times 10^{-17}$
- C$-1.96 \times 10^{-17},-2.18 \times 10^{-18}$
- D$-8.72 \times 10^{-17},-1.96 \times 10^{-17}$
Solution
Key Idea Use relation:
$$
E_n=-2.18 \times 10^{-18} \cdot \frac{Z^2}{n^2} \mathrm{~J} / \text { ion }
$$
where,
$E_n=$ energy of an electron in $n$th orbit
$Z=$ atomic number, $n=$ orbit number
For helium ion $\left(\mathrm{He}^{+}\right)$
$$
\begin{aligned}
Z & =2, n=1 \\
\therefore \quad E_n & =-2.18 \times 10^{-18} \times \frac{2^2}{1^2} \\
& =-8.72 \times 10^{-18} \mathrm{~J}
\end{aligned}
$$
For lithium ion $\left(\mathrm{Li}^{2+}\right)$
$$
\begin{aligned}
Z & =3, n=3 \\
\therefore \quad E_n & =-2.18 \times 10^{-18} \times \frac{3^2}{3^2} \\
& =-2.18 \times 10^{-18} \mathrm{~J}
\end{aligned}
$$
Hence, option (a) is the correct answer.
Hence, the correct answer is option: 'A'.