The radius of the second orbit of a hydrogen | vedclass

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Question

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

Vedclass · Accepted Answer

$4, Be^{3+}$

Vedclass · Answer

(A) The radius of an orbit in a hydrogen-like species is given by the formula $r_n = 0.529 	imes \frac{n^2}{Z} \ \mathring{A}$.
For the second orbit of a hydrogen atom $(H)$, $n_1 = 2$ and $Z_1 = 1$. So, $r_1 = 0.529 	imes \frac{2^2}{1} = 0.529 	imes 4$.
For the orbit $n$ of an ion $x$ with atomic number $Z_2$, $r_2 = 0.529 	imes \frac{n^2}{Z_2}$.
Given $r_1 = r_2$, we have $4 = \frac{n^2}{Z_2}$, which implies $n^2 = 4Z_2$.
Checking the options:
For $A$: $Be^{3+}$ has $Z=4$. $n^2 = 4 	imes 4 = 16$, so $n=4$. This matches.
For $B$: $Li^{2+}$ has $Z=3$. $n^2 = 4 	imes 3 = 12$ (not a perfect square).
For $C$: $Be^{2+}$ has $Z=4$. $n^2 = 16$, $n=4$. However, $Be^{2+}$ is not a hydrogen-like species (it has $2$ electrons).
For $D$: $He^{+}$ has $Z=2$. $n^2 = 4 	imes 2 = 8$ (not a perfect square).
Thus, the correct option is $A$.

The radius of the second orbit of a hydrogen atom is the same as that of orbit $n$ of an ion $x$. $n$ and $x$ are respectively:

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