The electron in the nth orbit of Li2+ is excited to (n+1) orbit using the radiation of energy 1.47×10−17J (as shown in the diagram). The value of n is ____________. Given: RH=2.18×10−18J

e−in nth orbit Jumped nth to (n+1)th orbit of Li+2 by absorppivity 1.47×10−17J En+1−En=13.6×Z21n2−1(n+1)2ev13.6×9×1.6×10−191n2−1(n+1)2=1.47×10−171n2−1(n+1)2=34∴n=1

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The electron in the n^th orbit of Li²⁺ is excited to (n+1) orbit using the radiation of energy $1.47 \times 10^{- 17} J$ (as shown in the diagram). The value of n is ____________.
Given: $R_{H} = 2.18 \times 10^{- 18} J$

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Detailed Solution

$e^{-} in n^{th} orbit Jumped n^{th} to (n + 1)^{th} orbit of {Li}^{+ 2} by absorppivity 1.47 \times 10^{- 17} J \begin{array}{l} E_{n + 1} - E_{n} = 13.6 \times Z^{2} (\frac{1}{n^{2}} - \frac{1}{(n + 1)^{2}}) ev \\ 13.6 \times 9 \times 1.6 \times 10^{- 19} (\frac{1}{n^{2}} - \frac{1}{(n + 1)^{2}}) = 1.47 \times 10^{- 17} \\ \frac{1}{n^{2}} - \frac{1}{(n + 1)^{2}} = \frac{3}{4} \\ ∴ n = 1 \end{array}$