AP EAMCET · PHYSICS · Atomic Physics
A hydrogen atom falls from \(\mathrm{n}^{\text {th }}\) higher energy orbit to first energy orbit ( \(\mathrm{n}=1\) ). The energy released is equal to 12.75 ev . The \(n^{\text {th }}\) orbit is
- A \(n=4\)
- B \(n=3\)
- C \(n=6\)
- D \(n=5\)
Answer & Solution
Correct Answer
(A) \(n=4\)
Step-by-step Solution
Detailed explanation
In hydrogen atom, \(\begin{aligned} & \Delta \mathrm{E}=13.6\left(\frac{1}{\mathrm{n}_1^2}-\frac{1}{\mathrm{n}_2^2}\right) \\ & \Rightarrow 12.75=13.6\left(\frac{1}{1^2}-\frac{1}{\mathrm{n}^2}\right) \\ & \therefore \mathrm{n}=4 .\end{aligned}\)
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