Question

The radius of Bohr's first orbit in a hydrogen atom is 0.053 nm. The radius of second orbit in $H{e}^{+}$ could be:

• A. $0.0265nm$
• B. $0.0530nm$
• C. $0.106nm$
• D. $0.2120nm$

Answer 1

The correct option is C $0.106nm$

$r_n=r_1\times \frac{n^2}{Z}$
$r_2=r_1\times \frac{2^2}{2}=0.106nm$