The radius of the fourth orbit in He^{+} ion | vedclass

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(C) The radius of an orbit in a hydrogen-like species is given by the formula: $r_n = 0.529 	imes \frac{n^2}{Z} \mathring{A}$.
Since $1 \mathring{A}

Vedclass · Accepted Answer

$264.50$

Vedclass · Answer

(C) The radius of an orbit in a hydrogen-like species is given by the formula: $r_n = 0.529 	imes \frac{n^2}{Z} \mathring{A}$.
Since $1 \mathring{A} = 100 \ pm$, the formula in $pm$ is $r_n = 52.9 	imes \frac{n^2}{Z} \ pm$.
For $He^{+}$ ion $(Z = 2)$, the radius of the fourth orbit $(n = 4)$ is:
$R_1 = 52.9 	imes \frac{4^2}{2} = 52.9 	imes \frac{16}{2} = 52.9 	imes 8 = 423.2 \ pm$.
For $Li^{2+}$ ion $(Z = 3)$, the radius of the third orbit $(n = 3)$ is:
$R_2 = 52.9 	imes \frac{3^2}{3} = 52.9 	imes 3 = 158.7 \ pm$.
Therefore, $(R_1 - R_2) = 423.2 - 158.7 = 264.5 \ pm$.

The radius of the fourth orbit in $He^{+}$ ion is '$R_1$' $pm$ and the radius of the third orbit in $Li^{2+}$ ion is '$R_2$' $pm$. The value of $(R_1 - R_2)$ in $pm$ is:

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