Radius of 1^{st} orbit of H and some orbit of | vedclass

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(A) The radius of the $n^{th}$ orbit is given by $r_n = 0.529 	imes \frac{n^2}{Z} \ \mathring{A}$.
For $H$ atom,$Z = 1$ and $n = 1$,so $r_H = 0.529

Vedclass · Accepted Answer

$-54.4$

Vedclass · Answer

(A) The radius of the $n^{th}$ orbit is given by $r_n = 0.529 	imes \frac{n^2}{Z} \ \mathring{A}$.
For $H$ atom,$Z = 1$ and $n = 1$,so $r_H = 0.529 	imes \frac{1^2}{1} = 0.529 \ \mathring{A}$.
For $Be^{3+}$ ion,$Z = 4$. Let the orbit number be $n$.
Given $r_H = r_{Be^{3+}}$,so $0.529 	imes \frac{1^2}{1} = 0.529 	imes \frac{n^2}{4}$.
$1 = \frac{n^2}{4} \implies n^2 = 4 \implies n = 2$.
The energy of an orbit is given by $E_n = -13.6 	imes \frac{Z^2}{n^2} \ 	ext{eV}$.
For $Be^{3+}$,$Z = 4$ and $n = 2$,so $E = -13.6 	imes \frac{4^2}{2^2} = -13.6 	imes \frac{16}{4} = -13.6 	imes 4 = -54.4 \ 	ext{eV}$.

Radius of $1^{st}$ orbit of $H$ and some orbit of $Be^{3+}$ is same. Energy of their orbit of $Be^{3+}$ is ............$eV$.

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