Question

In hydrogen atom an electron jumps from state \(i\) to \(f\). Radius of path of electron in state \(i\) and \(f\) are \(r_i\) and \(r_f\) respectively such that \[ \frac{r_i}{r_f}=\frac{4}{1} \] If Rydberg constant \(R = 1.0976\times10^7\ \text{m^{-1}\), find wavelength of emitted photon (in \(\text{\AA}\)). (\(r_i\) and \(r_f\) are minimum possible radius.)}

Answer 1

Correct Answer: 1215

Concept: For hydrogen atom, \[ r_n \propto n^2 \] Thus \[ \frac{r_i}{r_f}=\frac{n_i^2}{n_f^2} \] Step 1: Find quantum numbers \[ \frac{n_i^2}{n_f^2}=4 \] \[ \frac{n_i}{n_f}=2 \] Minimum possible integers: \[ n_i=2,\quad n_f=1 \] Step 2: Use Rydberg formula \[ \frac{1}{\lambda}=R\left(\frac{1}{n_f^2}-\frac{1}{n_i^2}\right) \] \[ \frac{1}{\lambda}=1.0976\times10^7\left(1-\frac14\right) \] \[ \frac{1}{\lambda}=1.0976\times10^7\left(\frac34\right) \] Step 3: Find wavelength \[ \lambda=\frac{1}{8.232\times10^6} \] \[ \lambda=1.215\times10^{-7}\ \text{m} \] \[ \lambda=1215\ \text{\AA} \]