What is the radius of the ground state of $\mathrm{He}^{+}$?
A) $0.0529 \mathrm{~nm}$
B) $0.212 \mathrm{~nm}$
C) $0.193 \mathrm{~nm}$
D) $0.0265 \mathrm{~nm}$
E) $0.106 \mathrm{~nm}$

Question

Quizplus · Accepted Answer

The radius of the ground state of a hydrogen-like ion is given by $ r_n = n^2 \times \frac{a_0}{Z} $, where $ a_0 = 0.0529 \mathrm{~nm} $ is the Bohr radius, $ n $ is the principal quantum number, and $ Z $ is the atomic number. For $\mathrm{He}^{+}$, $ n = 1 $ and $ Z = 2 $, so $ r_1 = \frac{0.0529 \mathrm{~nm}}{2} = 0.0265 \mathrm{~nm} $.

What Is the Radius of the Ground State Of He+\mathrm{He}^{+}He+ ? A)

What Is the Radius of the Ground State Of $\mathrm{He}^{+}$ ?
A)