Question

In the first excited state of hydrogen atom, the energy of its electron is 10.2 eV. The radial distance of the electron from the hydrogen nucleus in this case is approximately: ____.

• A. 2.1 $\times$ 10⁻¹¹ m
• B. 2.1 $\times$ 10⁻¹⁰ m
• C. 2.1 $\times$ 10⁻⁹ m
• D. 2.1 $\times$ 10⁻⁸ m

Hint:

Always remember: $n=1$ is Ground State, $n=2$ is 1st Excited State, $n=3$ is 2nd Excited State. Using the wrong $n$ is the most common mistake in these problems.

Answer 1

The Correct Option is B

Step 1: Understanding the Topic:
This problem is based on the "Bohr Model of the Atom." Bohr proposed that electrons move in specific circular orbits where their angular momentum is quantized. A significant part of this theory is the mathematical prediction of the radii of these allowed orbits.
Step 2: Key Formulas and Approach:
The radius of the $n$-th orbit for a Hydrogen atom ($Z=1$) is given by: \[ r_n = r_0 \times n^2 \] Where:
$r_0$ (Bohr radius) $\approx 0.529 \text{ \AA} = 0.529 \times 10^{-10} \text{ m}$.
$n$ is the principal quantum number.

Step 3: Detailed Explanation:

Identify the state: The problem specifies the "first excited state." In atomic physics:
Ground state: $n = 1$.
First excited state: $n = 2$.
Second excited state: $n = 3$.

Calculate the radius: Using $n = 2$: \[ r_2 = 0.529 \times 10^{-10} \text{ m} \times (2)^2 \] \[ r_2 = 0.529 \times 10^{-10} \times 4 \]
Compute the product: \[ r_2 = 2.116 \times 10^{-10} \text{ m} \]
The value $10.2 \text{ eV}$ mentioned in the question is the excitation energy (energy required to move from $n=1$ to $n=2$), which confirms we are indeed looking at the $n=2$ orbit.
Step 4: Final Answer:
The radial distance is approximately 2.1 $\times$ 10⁻¹⁰ m.