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

Options

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

Explanation

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

### Step 1: Understand the given information
- The radius of the second excited state (n = 3) is given as \( r = 4.65 \, \text{Å} \).
- We need to find the de-Broglie wavelength \( \lambda \) of the electron.

### Step 2: Use the formula for de-Broglie wavelength
The de-Broglie wavelength \( \lambda \) is given by the formula:
\[ \lambda = \frac{h}{mv} \]
where:
- \( h \) is Planck's constant (\( 6.626 \times 10^{-34} \, \text{Js} \)),
- \( m \) is the mass of the electron (\( 9.11 \times 10^{-31} \, \text{kg} \)),
- \( v \) is the velocity of the electron.

### Step 3: Find the velocity of the electron
From Bohr's model, the angular momentum of the electron in the nth orbit is quantized and given by:
\[ L = mvr = n\frac{h}{2\pi} \]
For the second excited state (n = 3):
\[ L = 3\frac{h}{2\pi} \]
Substituting \( r = 4.65 \times 10^{-10} \, \text{m} \):
\[ mv(4.65 \times 10^{-10}) = 3\frac{h}{2\pi} \]
Rearranging to find \( mv \):
\[ mv = \frac{3h}{2\pi (4.65 \times 10^{-10})} \]

### Step 4: Substitute the values
Substituting \( h = 6.626 \times 10^{-34} \, \text{Js} \):
\[ mv = \frac{3 \times 6.626 \times 10^{-34}}{2\pi (4.65 \times 10^{-10})} \]
Calculating this:
\[ mv = \frac{1.9878 \times 10^{-33}}{2\pi (4.65 \times 10^{-10})} \]
\[ mv = \frac{1.9878 \times 10^{-33}}{2.918 \times 10^{-9}} \approx 6.80 \times 10^{-25} \, \text{kg m/s} \]

### Step 5: Calculate the de-Broglie wavelength
Now substitute \( mv \) back into the de-Broglie wavelength formula:
\[ \lambda = \frac{h}{mv} = \frac{6.626 \times 10^{-34}}{6.80 \times 10^{-25}} \]
Calculating this:
\[ \lambda \approx 9.74 \times 10^{-10} \, \text{m} = 9.74 \, \text{Å} \]

### Final Answer
The de-Broglie wavelength of the electron in the second excited state is approximately \( 9.74 \, \text{Å} \).