The de-Broglie wavelength of electron in gound state of an hydrogen atom is

Options

A. `1.06 Å`
B. `1.52 Å`
C. `0.53 Å`
D. `3.33 Å`

Answer

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

### Step 1: Understand the relationship between de-Broglie wavelength and momentum
The de-Broglie wavelength (λ) 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 2: Use the Bohr model for the hydrogen atom
For the ground state of the hydrogen atom (n=1), we can use the Bohr model, where the angular momentum is quantized:
\[ mvr = \frac{nh}{2\pi} \]
For the ground state (\(n=1\)), this simplifies to:
\[ mvr = \frac{h}{2\pi} \]

### Step 3: Relate the radius of the orbit to the de-Broglie wavelength
From the equation \(mvr = \frac{h}{2\pi}\), we can express \(v\) in terms of \(r\):
\[ v = \frac{h}{2\pi mr} \]
Substituting this into the de-Broglie wavelength equation gives:
\[ \lambda = \frac{h}{m \left(\frac{h}{2\pi mr}\right)} = 2\pi r \]

### Step 4: Calculate the radius of the ground state orbit
The radius of the first orbit (ground state) for hydrogen is approximately:
\[ r = 0.53 \, \text{Å} = 0.53 \times 10^{-10} \, \text{m} \]

### Step 5: Calculate the de-Broglie wavelength
Now we can substitute the radius into the equation for the de-Broglie wavelength:
\[ \lambda = 2\pi r = 2 \times 3.14 \times (0.53 \times 10^{-10} \, \text{m}) \]
Calculating this gives:
\[ \lambda \approx 3.33 \times 10^{-10} \, \text{m} = 3.33 \, \text{Å} \]

### Final Answer
The de-Broglie wavelength of the electron in the ground state of a hydrogen atom is approximately:
\[ \lambda \approx 3.33 \, \text{Å} \]