Question

If $\left(\frac{0.51\times {10}^{-10}}{4}\right)m$, is the radius of smallest electron orbit in hydrogen like atom, then this atom is

• A. H-atom
• B. $H{e}^{+}$
• C. $L{i}^{2+}$
• D. $B{e}^{3+}$

Answer 1

Answer: (D)

$B{e}^{3+}$

Radius of hydrogen like atom
$\Rightarrow r_n=\frac{n^2}{Z}{r}_0$ ....(1)
where, $r_0=0.51\times {10}^{-10}m$
And $r_n=\frac{0.51\times {10}^{-10}}{4}m$ (Given) ....(2)
Equating (1) and (2), $\frac{0.51\times {10}^{-10}}{4}=$ $\frac{1^2}{Z}\times (0.51\times {10}^{-10})$

At ground state ( i.e. for smallest radius )$n=1$, therefore we get $Z=4$
Hence, it is Berilium $\left(B{e}^{3+}\right)$