Question

The radius of the ground orbit of electron in hydrogen-like atom is $\left(0.51\times {\left(10\right)}^{-10}/4\right)$ metre; then identify the atom?

• 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 nth orbit is
$r_n^z=\frac{n^2}{Z}{r}_0$
Where $r_0=0.51\times 10^{-10}$ metre
Here, $r_n^z=\frac{0.51\times 10^{-10}}{4}$ metre
In the ground state, $n=1$
$\therefore \frac{0.51\times 10^{-10}}{4}=\frac{1^2}{Z}\times 0.51\times 10^{-10}$
$Z=4$
So, the atom is triply ionised beryllium.