Question

The radius of hydrogen atom in ground state is $0.53\stackrel{\circ }{A}$. The radius of the first orbit of$L{i}^{2+}$ will be:

• A. $1.27\stackrel{\circ }{A}$
• B. $0.18\stackrel{\circ }{A}$
• C. $0.57\stackrel{\circ }{A}$
• D. $0.99\stackrel{\circ }{A}$

Answer 1

The correct option is B $0.18\stackrel{\circ }{A}$

Bohr's radius for $n^{th}$ orbit, $\left(r_n\right)=0.529\left[\frac{n^2}{Z}\right]\stackrel{\circ }{A}$
where $n=\text{Orbit number},Z=\text{Atomic number}$

For $L{i}^{2+}$: $(n=1),Z=3$

$\left(r_{L{i}^{2+}}\right)=\frac{r_H}{3}=\frac{0.53}{3}\stackrel{\circ }{A}=0.177\stackrel{\circ }{A}$