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 $\left( Be ^{3+}\right)$.