Ncert Exercises & Solutions

12.8 The radius of the innermost electron orbit of a hydrogen atom is 5.3×10^-11 m. What are the radii of the n = 2 and n =3 orbits?

Hint: Use the relation of the orbital radius with the orbital number.

Step 1: Find the the radius of the orbit for n = 2 and n = 3.

The radius of the innermost orbit of a hydrogen atom,

r_{1} = 5.3 \times 10^{- 11} m

Let

r_{2}

be the radius of the orbit at n = 2. It is related to the radius of the innermost orbit as:

r_{2} = {(n)}^{2} r_{1}

= 4 \times 5.3 \times 10^{- 11} = 2.12 \times 10^{- 10} m

For n = 3, we can write the corresponding electron radius as:

r_{3} = {(n)}^{2} r_{1}

= 9 \times 5.3 \times 10^{- 11} = 4.77 \times 10^{- 10} m

Hence, the radii of an electron for n=2 and n = 3 orbits are

2.12 \times 10^{- 10} m

and

4.77 \times 10^{- 10} m

respectively.