Question

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

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

Answer 1

Answer: (D)

For hydrogen like atom, the radius of $n$th orbit
$r_n^7=\frac{r^2}{Z}{a}_0$
Here, $a_0=0.51\times 10^{-10}m$
$\therefore r_n^2=\frac{0.51\times 10^{-10}}{4}m$
In the ground state, $n=1$
$\therefore \frac{0.51\times 10^{-10}}{4}=\frac{1^2}{Z}\times 0.51\times 10^{-10}$
$\therefore Z=4$
So, the atom is triply ionised beryllium $\left(B{e}^{3+}\right)$.