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Question

For hydrogen-like atom, the radius of ${n^{th}}$ orbit is

$r_n^Z = \frac{{{n^2}}}{Z}{r_0}$

Where $\quad \mathrm{r}_{0}=0.51 	imes 10^{-10}$ met

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$Be^{3+}$

Vedclass · Answer

For hydrogen-like atom, the radius of ${n^{th}}$ orbit is

$r_n^Z = \frac{{{n^2}}}{Z}{r_0}$

Where $\quad \mathrm{r}_{0}=0.51 	imes 10^{-10}$ metre

Here, $r_{n}^{z}=\frac{0.51 	imes 10^{-10}}{4} $ metre 

In the ground state, $n=1$

$	herefore \frac{0.51 	imes 10^{-10}}{4} =\frac{1^{2}}{Z} 	imes 0.51 	imes 10^{-10} $ 

$	herefore \quad Z =4 $

So, the atom is triply ionised beryllium.

The radius of the smallest electron orbit in hydrogen-like ion is $(0.51/4 \times 10^{-10})$ metre; then it is

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