COMEDK · Chemistry · 2. Structure of Atom
Bohr's radius of 2 nd orbit of \(\mathrm{Be}^{3+}\) is equal to that of
- A 1st orbit of hydrogen
- B 4th orbit of hydrogen
- C \(3 \mathrm{rd}\) obit of \(\mathrm{Li}^{2+}\)
- D 2nd orbit of \(\mathrm{He}^{+}\)
Answer & Solution
Correct Answer
(A) 1st orbit of hydrogen
Step-by-step Solution
Detailed explanation
Correct option is A.
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