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JEE Mains · Physics · STD 12 - 12. atoms

In a hydrogen atom the electron makes a transition from \(( n +1)^{\text {th }}\) level to the \(n ^{\text {th }}\) level. If \(n>>1,\) the frequency of radiation emitted is proportional to

  1. A \(\frac{1}{n^{4}}\)
  2. B \(\frac{1}{n^{3}}\)
  3. C \(\frac{1}{n^{2}}\)
  4. D \(\frac{1}{n}\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(\frac{1}{n^{3}}\)

Step-by-step Solution

Detailed explanation

In hydrogen atom, \(E _{ n }=\frac{- E _{0}}{ n ^{2}}\) Where \(E _{0}\) is Ionisation Energy of \(H\). \(\rightarrow\) For transition from \((n+1)\) to \(n,\) the energy of emitted radiation is equal to the difference in energies of levels. \(\Delta E = E _{ n +1}- E _{ n }\)…
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