The electron in the n^{th} orbit of Li^{2+} i | vedclass

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(C) The energy difference for the transition is given by the formula: $\Delta E = R_H Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$.For $Li^{2

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(C) The energy difference for the transition is given by the formula: $\Delta E = R_H Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} ight)$.For $Li^{2+}$,the atomic number $Z = 3$.Substituting the values: $1.47 	imes 10^{-17} = 2.18 	imes 10^{-18} 	imes 3^2 \left( \frac{1}{n^2} - \frac{1}{(n+1)^2} ight)$.$1.47 	imes 10^{-17} = 1.962 	imes 10^{-17} \left( \frac{(n+1)^2 - n^2}{n^2(n+1)^2} ight)$.$\frac{1.47}{1.962} \approx 0.75 = \frac{3}{4} = \frac{2n+1}{n^2(n+1)^2}$.For $n = 1$: $\frac{2(1)+1}{1^2(2^2)} = \frac{3}{4}$.Thus,$n = 1$.

The electron in the $n^{th}$ orbit of $Li^{2+}$ is excited to $(n+1)$ orbit using the radiation of energy $1.47 \times 10^{-17} \ J$. The value of $n$ is $....$. Given $R_H = 2.18 \times 10^{-18} \ J$.

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