A collection of H-atoms in 9th excited state returns to ground state. Calculate ratio of total number of spectral lines emitted without emitting any line in Brackett series to number of Brackett series lines.

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$\frac{39}{6}$

$\frac{45}{6}$

$\frac{45}{39}$

answer is D.

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Detailed Solution

We have given that;

$n = 10 to n = 1$

Total lines $= \frac{10 \times 9}{2} = 45$ ; No of Brackett lines

We get also,

$\begin{array}{l} 4 \to 3 \\ 4 \to 2 \\ 4 \to 1 \end{array}$ $\begin{array}{r} 10 \to 4 \\ 9 \to 4 \\ 8 \to 4 \\ 7 \to 4 \\ 6 \to 4 \\ 5 \to 4 \end{array}\} 6$

should not be considered

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