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. $He^{+}$
• C. $Li^{2 +}$
• D. $Be^{3 +}$

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.