The electron in the n th orbit of Li 2 + is excited to ( n + 1 ) orbit using the radiation of energy 1 . 47 × 10 - 17 J (as shown in the diagram). The value of n is ________ . Given: R H = 2 . 18 × 10 - 18 J [

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Solution

Q.

The electron in the $n^{th}$ orbit of ${Li}^{2 +}$ 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: $R_{H} = 2.18 \times 10^{- 18} J$ [2023]

Ans.
(1)

$Δ E = R_{H} Z^{2} (\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}})$

$or 1.47 \times 10^{- 17} = 2.18 \times 10^{- 18} \times 9 (\frac{1}{n^{2}} - \frac{1}{{(n + 1)}^{2}})$

$\Rightarrow \frac{1.47}{1.96} = \frac{3}{4} = \frac{1}{n^{2}} - \frac{1}{{(n + 1)}^{2}}$

On solving for $n$ , we get

$3 n^{4} + 6 n^{3} + 3 n^{2} - 8 n - 4 = 0$

If we take $n = 1$

$3 + 6 + 3 - 8 - 4 = 0$

$∴$ $n = 1$

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