Q. The radius of the Bohr orbit in the ground state of hydrogen atom is $0.5\ \mathring{\mathrm{A}}$. The radius of the orbit of the electron in the third excited state of $He^{+}$ will be
A) $8\ \mathring{\mathrm{A}}$
B) $4\ \mathring{\mathrm{A}}$
C) $0.5\ \mathring{\mathrm{A}}$
D) $0.25\ \mathring{\mathrm{A}}$
Explanation
(b) By using $r_{n} = r_{0}\frac{n^{2}}{Z}$; where $r_{0} =$ Radius of the Bohr orbit in the ground state atom. So for $He^{+}$ third excited state $n = 4,Z = 2,r_{0} = 0.5\mathring{\mathrm{A}} \Rightarrow r_{4} = 0.5 \times \frac{4^{2}}{2} = 4\mathring{\mathrm{A}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.