AP EAMCET · Chemistry · Structure of Atom
The angular momentum of an electron in a stationary state of \(\mathrm{Li}^{2+}(\mathrm{Z}=3)\) is \(\frac{3 h}{\pi}\). The radius and energy of that stationary state are respectively
- A \(3.174 Å,-5.45 \times 10^{-19} \mathrm{~J}\)
- B \(6.348 Å,-5.45 \times 10^{-19} \mathrm{~J}\)
- C \(6.348 Å,+5.45 \times 10^{-18} \mathrm{~J}\)
- D \(2.116 Å,-5.45 \times 10^{-19} \mathrm{~J}\)
Answer & Solution
Correct Answer
(B) \(6.348 Å,-5.45 \times 10^{-19} \mathrm{~J}\)
Step-by-step Solution
Detailed explanation
Angular momentum \(=\frac{3 \mathrm{~h}}{\pi} \Rightarrow \mathrm{~m}_{\mathrm{e}} \mathrm{vr}= \frac{\mathrm{nh}}{2 \pi}\) \(\frac{3 \mathrm{~h}}{\pi}=\mathrm{n} \cdot \frac{\mathrm{h}}{2 \pi} \Rightarrow \mathrm{n}=6\) For the Radius…
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