If an electron in a hydrogen atom is in its `2^("nd")` excited state, orbiting the nucleus in a stationary orbit or radius `4.65Å`, then its de - Broglie wavelength is

Options

A. `12.9Å`
B. `3.5Å`
C. `9.7Å`
D. `6.67Å`

Answer

To find the de Broglie wavelength of an electron in the second excited state of a hydrogen atom, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the given values**:
- The radius of the orbit, \( r = 4.65 \, \text{Å} = 4.65 \times 10^{-10} \, \text{m} \)
- The principal quantum number for the second excited state, \( n = 3 \)

2. **Use the de Broglie wavelength formula**:
According to de Broglie's hypothesis, the circumference of the electron's orbit is an integral multiple of the wavelength:
\[ 2 \pi r = n \lambda \]
where \( \lambda \) is the de Broglie wavelength.

3. **Rearranging the formula to find \( \lambda \)**:
\[ \lambda = \frac{2 \pi r}{n} \]

4. **Substituting the known values**:
\[ \lambda = \frac{2 \pi (4.65 \times 10^{-10} \, \text{m})}{3} \]

5. **Calculating the circumference**:
First, calculate \( 2 \pi (4.65 \times 10^{-10}) \):
\[ 2 \pi \approx 6.2832 \]
\[ 2 \pi (4.65 \times 10^{-10}) \approx 6.2832 \times 4.65 \times 10^{-10} \approx 2.94 \times 10^{-9} \, \text{m} \]

6. **Now divide by \( n \)**:
\[ \lambda = \frac{2.94 \times 10^{-9}}{3} \approx 9.8 \times 10^{-10} \, \text{m} \]

7. **Convert to angstroms**:
Since \( 1 \, \text{Å} = 10^{-10} \, \text{m} \):
\[ \lambda \approx 9.8 \, \text{Å} \]

### Final Answer:
The de Broglie wavelength of the electron in the second excited state of a hydrogen atom is approximately \( 9.8 \, \text{Å} \).

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