COMEDK · Chemistry · 2. Structure of Atom
Bohr's radius of \(2 \mathrm{nd}\) orbit of \(\mathrm{Be}^{3+}\) is equal to that of
- A 4th orbit of hydrogen
- B 2nd orbit of \(\mathrm{He}^{+}\)
- C 3 rd obit of \(\mathrm{Li}^{2+}\)
- D 1st orbit of hydrogen
Answer & Solution
Correct Answer
(D) 1st orbit of hydrogen
Step-by-step Solution
Detailed explanation
Bohr radius for \(n\)th orbit \(=0.53 Å \times \frac{n^2}{Z}\) where, \(Z=\) atomic number \(\therefore\) Bohr radius of 2 nd orbit of \(\mathrm{Be}^{3+}=\frac{0.53 \times(2)^3}{4}=0.53 Å\) For option (d) Bohr radius of 1st orbit of…
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