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?

Hint :The electron undergoes excitation within the energy levels. The radius of the innermost electron orbit means the radius of the first orbit which takes the value of $ n = 1 $ . By knowing the value of radius of the first orbit the radius of higher orbits can be calculated from the orbit number.

Complete Step By Step Answer:
Hydrogen is the atom with only one electron in its shell. This one electron undergoes excitation within the energy levels or orbits. They are different energy levels like electronic, vibrational, rotational and translational energy levels.
The radius of the orbit will be given as:
 $ {r_n} = \dfrac{{{n^2}{h^2}}}{{4{\pi ^2}mZ{e^2}}} $
r is the radius of orbit
n is the number of orbit from which transition occurs
h is Planck’s constant
m is the mass of orbit
Z is atomic number
e is charge on proton
All the above values are constant except the value of orbit number represented by n.
Given that the radius of the innermost electron orbit of a hydrogen atom is $ 5.3 \times {10^{11}}m $ which means
 $ {r_1} = \dfrac{{{{\left( 1 \right)}^2}{h^2}}}{{4{\pi ^2}mZ{e^2}}} = 5.3 \times {10^{11}}m $
Now, the radii of the $ n = 2 $ and $ n = 3 $ orbits can be determined by substituting the value of $ 2 $ and $ 3 $ in the value of $ 1 $ .
The radius of orbit $ n = 2 $ is $ {r_2} = \dfrac{{{{\left( 2 \right)}^2}{h^2}}}{{4{\pi ^2}mZ{e^2}}} = 4 \times 5.3 \times {10^{11}}m = 21.2 \times {10^{11}}m $
The radius of orbit $ n = 3 $ is $ {r_3} = \dfrac{{{{\left( 3 \right)}^2}{h^2}}}{{4{\pi ^2}mZ{e^2}}} = 9 \times 5.3 \times {10^{11}}m = 47.7 \times {10^{11}}m $ .

Note :
The radius of inner orbit means the radius of the first orbit. As the inner means the lowest or ground state. Thus, the radii of the orbits can be known by knowing the orbit number. Scientist Bohr successfully discovered the method only for the atoms with one electron.