Question

Electrons in a sample of H-atoms make transition from state n=x to some lower excited state. The emmission spectrum from the sample is found to contain only the lines belonging to particular series. If one of the photons had an energy of 0.6375 eV. Then find the value of x. [Take 0.63756 eV= $\frac{3}{4}\times$ 0.85 eV]

Answer 1

We have

$\Delta E=\frac{3}{4}\times 0.85eV$

as energy=0.6375 the photon will belong to Brackett series (as for Brackett ($0.31\le E\le 0.85$)

$0.85\times (1-\frac{1}{4})=13.6(\frac{1}{4^2}-\frac{1}{n^2})$

$0.85\times (1-\frac{1}{4})=\frac{13.6}{16}[1-(\frac{4}{n}{)}^2]$

$\therefore \frac{4}{n}=\frac{1}{2}⟹n=8$

Hence x=8