Question

In a hydrogen atom, the radius of the $n^{th}$ Bohr orbit is proportional to

• A. $n$
• B. $1/n$
• C. $n^{2}$
• D. $1/n^{2}$
• E. $n^{3}$

Hint:

The first orbit ($n=1$) has a radius of approximately $0.529$~\AA. Since it follows $n^2$, the second orbit is 4 times larger ($2.12$~\AA) and the third is 9 times larger ($4.76$~\AA).

Answer 1

The Correct Option is C

Concept:
Physics - Atomic Structure (Bohr Model).
Step 1: Identify the formula for Bohr radius.
According to Bohr's model, the radius $r_{n}$ of the $n^{th}$ orbit of a hydrogen-like atom is: $$ r_{n} = \frac{\epsilon_{0} h^{2} n^{2}}{\pi m e^{2} Z} $$ For a hydrogen atom, $Z = 1$.
Step 2: Analyze the constants and variables.
Everything in the formula except $n$ (and $Z$ for different atoms) is a constant.
Step 3: Determine proportionality.
From the formula: $$ r_{n} \propto n^{2} $$
Step 4: Conclusion.
The radius of the orbit increases with the square of the principal quantum number.