The radius of the fifth orbit of the Li^{++} | vedclass

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(D) The radius of the $n^{th}$ orbit for a hydrogen-like atom is given by the formula: $r_n = r_0 	imes \frac{n^2}{Z}$, where $r_0$ is the Bohr radiu

Vedclass · Accepted Answer

$425$

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(D) The radius of the $n^{th}$ orbit for a hydrogen-like atom is given by the formula: $r_n = r_0 	imes \frac{n^2}{Z}$, where $r_0$ is the Bohr radius $(0.51 \ \mathring{A})$, $n$ is the orbit number, and $Z$ is the atomic number.For $Li^{++}$ (Lithium ion), the atomic number $Z = 3$.The orbit number $n = 5$.Substituting the values: $r_5 = 0.51 \ \mathring{A} 	imes \frac{5^2}{3} = 0.51 	imes \frac{25}{3} \ \mathring{A}$.$r_5 = 0.51 	imes 8.333 = 4.25 \ \mathring{A}$.Since $1 \ \mathring{A} = 10^{-10} \ m$, we have $r_5 = 4.25 	imes 10^{-10} \ m$.To express this in terms of $10^{-12} \ m$, we multiply by $100/100$: $r_5 = 425 	imes 10^{-12} \ m$.Therefore, the missing value is $425$.

The radius of the fifth orbit of the $Li^{++}$ ion is $......... \times 10^{-12} \ m$. Take: radius of the hydrogen atom (first Bohr radius) $r_0 = 0.51 \ \mathring{A}$.

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