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COMEDK · Physics · 28. Atomic Physics

The electron in a hydrogen atom makes a transition from \(n=n_{1}\) to \(n=n_{2}\) state. The time period of the electron in the initial state \(n_{1}\) is eight times that in the final state \(n_{2}\). The possible values of \(n_{1}\) and \(n_{2}\) are

  1. A \(n_{1}=8, n_{2}=1\)
  2. B \(n_{1}=4, n_{2}=2\)
  3. C \(n_{1}=2, n_{2}=4\)
  4. D \(n_{1}=1, n_{2}=8\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(n_{1}=4, n_{2}=2\)

Step-by-step Solution

Detailed explanation

Here, for hydrogen atom, \(Z=1\) We know that, time period of electron in \(n\)th orbit is given as \(\begin{aligned} T_{n} & \propto n^{3} \\ \therefore \quad & \frac{T_{n_{1}}}{T_{n_{2}}}=\frac{n_{1}^{3}}{n_{2}^{3}} \end{aligned}\) Since, \(\quad T_{n_{1}}=8 T_{n_{2}}\)…
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