Question

The orbit having Bohr radius equal to $1^{st}$ Bohr orbit of $H-$atom is:

• A. $n=2$ of $H{e}^{+}$
• B. $n=2$ of $B^{+4}$
• C. $n=3$ of $L{i}^{+2}$
• D. $n=2$ of $B{e}^{+3}$

Answer 1

The correct option is D $n=2$ of $B{e}^{+3}$

r$=\frac{n^2{h}^2}{4{\pi }^{2}m{e}^2}\times \frac{1}{z}$

=$0.529\times \frac{n^2}{z}$

where n=Orbit number

z=atomic number

n=1 of H atom

r$=\frac{1}{1}=1$

For $n=2$ of $B^{+4}$

r$=\frac{4}{5}$

For $n=2$ of ${He}^{+}$

r$=\frac{4}{2}=2$

For $n=3$ of ${Li}^{+2}$

r$=\frac{9}{3}=3$

For $n=2$ of ${Be}^{+2}$

r$=\frac{4}{4}=1$

So radius of 1 st orbit of H atom is equal to $n=2$ of ${Be}^{+3}$ atom