JEE Mains · Physics · STD 12 - 12. atoms
In the hydrogen atom, the electron makes a transition from the higher orbit (\(i\)) to a lower orbit (\(f\)). The ratio of the radius of the orbits in given by \(r_i : r_f = 16 : 4\). The wavelength of photon emitted due to this transition is _____ nm.
(Given Rydberg constant \(= 1.0973 \times 10^7\) /m)
- A \(121\)
- B \(242\)
- C \(486\)
- D \(974\)
Answer & Solution
Correct Answer
(C) \(486\)
Step-by-step Solution
Detailed explanation
The radius of the \(n\)-th orbit in a hydrogen atom is proportional to the square of the principal quantum number \(n\), i.e., \(r_n \propto n^2\). Given the ratio of the radii of the initial and final orbits: \(\dfrac{r_i}{r_f} = \dfrac{16}{4}\) This implies:…
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