Question:medium

The Bohr orbit radius for the hydrogen atom (n = 1) is approximately 0.530 \AA. The radius for the first excited state (n = 2) orbit is (in \AA)

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Always remember the sequence of squared integers for Bohr radii: 1, 4, 9, 16... Just multiply the base ground state radius by the square of the principal quantum number. \
Updated On: Apr 22, 2026
  • 0.13
  • 1.06
  • 4.77
  • 2.12
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The Correct Option is D

Solution and Explanation

Step 1: Understanding the Concept:
According to Bohr's model of the atom, electrons revolve around the nucleus in specific circular orbits. The radius of these orbits is quantized and depends on the principal quantum number ($n$).
Step 2: Key Formula or Approach:
The radius of the $n^{th}$ orbit for a hydrogen-like atom is given by: \[ r_n = r_0 \times \frac{n^2}{Z} \] where $r_0$ is the Bohr radius (radius of the first orbit of Hydrogen), $n$ is the principal quantum number, and $Z$ is the atomic number.
Step 3: Detailed Explanation:
1. For Hydrogen, $Z = 1$. The radius of the first orbit ($n=1$) is given as $r_1 = 0.530$ Å.
2. The "first excited state" corresponds to $n = 2$.
3. Using the proportionality $r_n \propto n^2$: \[ r_2 = r_1 \times (2)^2 \] \[ r_2 = 0.530 \times 4 \]
4. Calculation: \[ r_2 = 2.120 \, \text{Å} \]
Step 4: Final Answer
The radius for the first excited state orbit is 2.12 Å.
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