Consider an electron in a hydrogen atom, revolving in its second excited state (having radius `4.65 Å`). The de-Broglie wavelength of this electron is :

Options

A. `3.5 Å`
B. `6.6 Å`
C. `12.9 Å`
D. `9.7 Å`

Answer

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

1. **Identify the Quantum Number**: The second excited state corresponds to the principal quantum number \( n = 3 \).

2. **Use the Formula for Circumference**: The electron revolves around the nucleus, and the circumference of its orbit is given by:
\[ 2\pi r = n \lambda \]
where \( r \) is the radius of the orbit, \( n \) is the principal quantum number, and \( \lambda \) is the de-Broglie wavelength.

3. **Substitute the Known Values**: We know the radius \( r = 4.65 \, \text{Å} = 4.65 \times 10^{-10} \, \text{m} \) and \( n = 3 \). Substituting these values into the equation gives:
\[ 2\pi (4.65 \times 10^{-10}) = 3 \lambda \]

4. **Rearranging the Equation**: We can rearrange the equation to solve for \( \lambda \):
\[ \lambda = \frac{2\pi (4.65 \times 10^{-10})}{3} \]

5. **Calculate \( \lambda \)**: Now we can calculate \( \lambda \):
\[ \lambda = \frac{2 \times 3.14 \times (4.65 \times 10^{-10})}{3} \]
\[ \lambda = \frac{6.28 \times 4.65 \times 10^{-10}}{3} \]
\[ \lambda = \frac{29.22 \times 10^{-10}}{3} \approx 9.74 \times 10^{-10} \, \text{m} \]

6. **Convert to Angstroms**: Since \( 1 \, \text{Å} = 10^{-10} \, \text{m} \), we can express the wavelength in angstroms:
\[ \lambda \approx 9.74 \, \text{Å} \]

Thus, the de-Broglie wavelength of the electron in the second excited state of a hydrogen atom is approximately \( 9.74 \, \text{Å} \).