Question

The ratio of radii of \(3^{rd}\) and \(6^{th}\) Bohr's orbit in a hydrogen atom is

• A. \(0.25\)
• B. \(0.33\)
• C. \(4\)
• D. \(3\)

Hint:

In Bohr’s model of hydrogen atom, \[ r_n \propto n^2 \] Hence, ratios of orbital radii can be found directly using the squares of principal quantum numbers.

Answer 1

The Correct Option is A

Step 1: Recall Bohr's formula for orbital radius.
In Bohr's model, the radius of the $n$-th orbit of a hydrogen atom is:
\[ r_n = n^2 a_0 \] where $a_0 = 0.529\,\text{\AA}$ is the Bohr radius. The key fact is $r_n \propto n^2$.

Step 2: Write the radius for the 3rd orbit.
\[ r_3 = 3^2 \times a_0 = 9a_0 \]

Step 3: Write the radius for the 6th orbit.
\[ r_6 = 6^2 \times a_0 = 36a_0 \]

Step 4: Form the ratio.
\[ \frac{r_3}{r_6} = \frac{9a_0}{36a_0} = \frac{9}{36} = \frac{1}{4} \]

Step 5: Express as a decimal.
\[ \frac{r_3}{r_6} = \frac{1}{4} = 0.25 \]

Step 6: State the answer.
The 3rd orbit is 4 times smaller than the 6th orbit, because radius scales as $n^2$.
\[ \boxed{0.25} \]