AP EAMCET · Chemistry · Structure of Atom
The difference between the radii of \(3^{\text {rd }}\) and \(2^{\text {nd }}\) orbit of H -atom is \(x\) \(\mathrm{pm}\). The difference between the radii of \(4^{\text {th }}\) and \(3^{\text {rd }}\) orbit of \(\mathrm{Li}^{2+}\) ion is \(y \mathrm{pm}. y: x\) is equal to
- A \(15: 7\)
- B \(7: 15\)
- C \(3: 1\)
- D \(1: 3\)
Answer & Solution
Correct Answer
(B) \(7: 15\)
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