JEE Mains · Physics · STD 12 - 12. atoms
The ratio of momentum of the photons of the \(1^{st}\) and \(2^{nd}\) line of Balmer series of Hydrogen atoms is \(\alpha/\beta\). The possible values of \(\alpha\) and \(\beta\) are:-
- A \(27\) and \(20\)
- B \(3\) and \(16\)
- C \(5\) and \(36\)
- D \(20\) and \(27\)
Answer & Solution
Correct Answer
(D) \(20\) and \(27\)
Step-by-step Solution
Detailed explanation
The momentum of a photon is given by \(p = \dfrac{h}{\lambda}\). From Rydberg's formula, the wave number is \(\dfrac{1}{\lambda} = R \left( \dfrac{1}{n_1^2} - \dfrac{1}{n_2^2} \right)\). For the first line of the Balmer series, the transition is from \(n_2 = 3\) to \(n_1 = 2\).…
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