Question

If the radius of the first orbit of hydrogen atom is $5.29\times {10}^{-11}$ meter, the radius of the seconds orbit for $H{e}^{+}$ will be-

• A. $21.16\times {10}^{-11}m$
• B. $15.87\times {10}^{-11}m$
• C. $10.58\times {10}^{-11}m$
• D. $2.64\times {10}^{-11}m$

Answer 1

The correct option is C $10.58\times {10}^{-11}m$

Given that $a_0=5.29\times {10}^{-11}meter$

we know that for any hydrogen atom like system $r_n=\frac{a_0{n}^2}{Z}$

so for $H{e}^{+}$ i.e $Z=2$ and its second orbit i.e $n=2$ so $r_2=\frac{4a_0}{2}=2a_0=2\times 5.29\times {10}^{-11}m=10.58\times {10}^{-11}m$