Radius of smallest electron orbit in hydrogen-like ion is (0.51/4 × 10^{-10}) metre; then it is

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12.Atoms

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

A

hydrogen atom

B

$H^+$

C

$Li^{2+}$

D

$Be^{3+}$

Solution

For hydrogen-like atom, the radius of ${n^{th}}$ orbit is

$r_n^Z = \frac{{{n^2}}}{Z}{r_0}$

Where $\quad \mathrm{r}_{0}=0.51 \times 10^{-10}$ metre

Here, $r_{n}^{z}=\frac{0.51 \times 10^{-10}}{4} $ metre

In the ground state, $n=1$

$\therefore \frac{0.51 \times 10^{-10}}{4} =\frac{1^{2}}{Z} \times 0.51 \times 10^{-10} $

$\therefore \quad Z =4 $

So, the atom is triply ionised beryllium.

Standard 12

Physics

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