The radius of the Bohr orbit in the ground state of hydrogen atom is 0.5 Å. The radius of the orbit of the electron in the third excited state of He+ will be
Answer & explanation
Correct answer: option 2
By using $r_n=r_0 \frac{n^2}{Z}$; Where $r_o=$ Radius of the Bohr orbit in the ground state atom. So for $He^{+}$ third excited state $n=4, Z=2, r_0=0.5 ~Å \Rightarrow r_4=0.5 \times \frac{4^2}{2}=4 ~Å$