Q71 / 90 · JEE Main 2012 (07 May Online) · Physics

The electron of a hydrogen atom makes a transition from the $(n+1)^{\text {th }}$ orbit to the $n^{\text {th }}$ orbit. For large $\mathrm{n}$ the wavelength of the emitted radiation is proportional to

  • A
    $n$
  • Correct answer
    B
    $n^3$
  • C
    $n^4$
  • D
    $n^2$

Solution

If $n_1=n$ and $n_2=n+1$ Maximum wavelength $\lambda_{\max }=\frac{n^2\left(n+1^2\right)}{(2 n+1 R)}$ Therefore, for large $\mathrm{n}, \lambda_{\max } \propto n^3$

Hence, the correct answer is option: 'B'.

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