Question

The most probable radius (in pm) for finding the electron in $H{e}^{+}$ is:

• A. $0.0$
• B. $52.9$
• C. $26.45$
• D. $105.8$

Answer 1

The correct option is C $26.45$

Bohr's radius$\left(r\right)$

$=\frac{0.529\times {10}^{-10}{n}^2}{Z}m=\frac{52.9n^2}{Z}pm$

Here, $n$(No. of shell)$=1$

Z=Atomic Number$=2$

$\therefore \cdot r=\frac{52.9\times 1^2}{2}=26.45$pm

Hence, the correct option is $C$