KCET · Physics · Atomic Physics
The ratio of area of first excited state to ground state of orbit of hydrogen atom'is
- A \(1: 16\)
- B \(1: 4\)
- C \(4: 1\)
- D \(16: 1\)
Answer & Solution
Correct Answer
(D) \(16: 1\)
Step-by-step Solution
Detailed explanation
Since, \(r_n \propto n^2\)
\(\therefore\) Area, \(A_n \propto r_n^2 \propto n^4\)
\(\Rightarrow \quad \frac{A_{n_1}}{A_{n_2}}=\left(\frac{n_1}{n_2}\right)^4=\left(\frac{2 n}{n}\right)^4=16\)
\(\therefore \quad A_{n_1}: A_{n_2}=16: 1\)
\(\therefore\) Area, \(A_n \propto r_n^2 \propto n^4\)
\(\Rightarrow \quad \frac{A_{n_1}}{A_{n_2}}=\left(\frac{n_1}{n_2}\right)^4=\left(\frac{2 n}{n}\right)^4=16\)
\(\therefore \quad A_{n_1}: A_{n_2}=16: 1\)
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