JEE Mains · Chemistry · STD 11 - 2. structure of atom
The electron in the \(n ^{\text {th }}\) orbit of \(Li ^{2+}\) is excited to \((n+1)\) orbit using the radiation of energy \(1.47 \times 10^{-17}\,J\) (as shown in the diagram). The value of \(n\) is \(....\).Given \(R_H=2.18 \times 10^{-18}\,J\)
- A \(2\)
- B \(3\)
- C \(1\)
- D \(4\)
Answer & Solution
Correct Answer
(C) \(1\)
Step-by-step Solution
Detailed explanation
\(\Delta E = R _{ H } Z ^2\left(\frac{1}{ n _1^2}-\frac{1}{ n _2^2}\right)\) \(1.47 \times 10^{-17}=2.18 \times 10^{-15} \times 9\left(\frac{1}{n^2}-\frac{1}{(n+1)^2}\right)\) \(\frac{1.47}{1.96}=\frac{3}{4}=\frac{1}{n^2}-\frac{1}{(n+1)^2}\) So, \(n =1\)
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