Question

The radius of the smaller electron orbit in hydrogen-like ion is $\frac{0.51\times {10}^{-10}}{4}$m, then it is?

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

Answer 1

The correct option is D $B{e}^{3+}$

For hydrogen like atom, the radius of nth orbit
$r_n^z=\frac{n^2}{Z}{a}_0$
Here, $a_0=0.51\times {10}^{-10}$m
$\therefore r_n^z=\frac{0.51\times {10}^{-10}}{4}$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$\left(B{e}^{3+}\right)$.