The radius of the innermost electron orbit of a hydrogen atom is $5.3 \times 10^{-11} \;m$. What are the radii of the $n=2$ and $n=3$ orbits?

(N/A) The radius of the innermost orbit of a hydrogen atom is given by $r_{1} = 5.3 \times 10^{-11} \;m$.
The radius of an orbit for a hydrogen atom is given by the formula $r_{n} = n^{2} r_{1}$,where $n$ is the principal quantum number.
For $n=2$:
$r_{2} = (2)^{2} \times r_{1} = 4 \times 5.3 \times 10^{-11} = 2.12 \times 10^{-10} \;m$.
For $n=3$:
$r_{3} = (3)^{2} \times r_{1} = 9 \times 5.3 \times 10^{-11} = 4.77 \times 10^{-10} \;m$.
Thus,the radii for $n=2$ and $n=3$ orbits are $2.12 \times 10^{-10} \;m$ and $4.77 \times 10^{-10} \;m$ respectively.