AP EAMCET · Chemistry · Structure of Atom
The energy and radius of electron present in second orbit of \(\mathrm{He}^{+}\)respectively are
- A \(-1.09 \times 10^{-18} \mathrm{~J}, 105.8 \mathrm{pm}\)
- B \(-8.72 \times 10^{-18} \mathrm{~J}, 211.6 \mathrm{pm}\)
- C \(-4.36 \times 10^{-18} \mathrm{~J}, 52.9 \mathrm{pm}\)
- D \(-2.18 \times 10^{-18} \mathrm{~J}, 105.8 \mathrm{pm}\)
Answer & Solution
Correct Answer
(D) \(-2.18 \times 10^{-18} \mathrm{~J}, 105.8 \mathrm{pm}\)
Step-by-step Solution
Detailed explanation
\begin{aligned} & \text { For } \mathrm{He}^{+}: Z=2, n=2 \\ & E_n=-2.18 \times 10^{-18} \cdot \frac{Z^2}{n^2} \cdot \mathrm{J} \\ & \therefore \quad E_n=-2.18 \times 10^{-18} \times \frac{4}{4} \\ & =-2.18 \times 10^{-18} \mathrm{~J} \\ & r_n=52.9 \times \frac{n^2}{Z}…
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