Question

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

• A. $1.59 \times 10^{-10}$ m
• B. $1.06 \times 10^{-10}$ m
• C. $1.43 \times 10^{-9}$ m
• D. $4.77 \times 10^{-10}$ m

Hint:

Since $r \propto n^2$, if you move from the 1st orbit to the 2nd, the radius increases \textbf{4 times}. If you move to the 3rd, it increases \textbf{9 times}!

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
According to Bohr's model, the radius of the $n^{th}$ orbit of a hydrogen atom is proportional to the square of the principal quantum number ($n^2$).
Step 2: Calculation:
Formula: $r_n = n^2 \times r_1$ Given: $r_1 = 5.3 \times 10^{-11}\text{ m}$ and $n = 3$. $$r_3 = 3^2 \times (5.3 \times 10^{-11})$$ $$r_3 = 9 \times 5.3 \times 10^{-11}$$ $$r_3 = 47.7 \times 10^{-11}\text{ m} = 4.77 \times 10^{-10}\text{ m}$$
Step 3: Final Answer:
The radius is $4.77 \times 10^{-10}$ m.