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

Options

A. `pir^(2)`
B. `2 pi r`
C. `pi r `
D. `sqrt(2pi r)`

Explanation

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

### Step-by-Step Solution:

1. **Understand the de-Broglie Wavelength Formula**:
The de-Broglie wavelength (\( \lambda \)) is given by the formula:
\[ \lambda = \frac{h}{p} \]
where \( h \) is Planck's constant and \( p \) is the momentum of the electron.

2. **Determine the Momentum of the Electron**:
According to Bohr's model of the hydrogen atom, the angular momentum (\( L \)) of the electron in a hydrogen atom is quantized and given by:
\[ L = n \frac{h}{2\pi} \]
where \( n \) is the principal quantum number. For the ground state, \( n = 1 \).

3. **Relate Angular Momentum to Momentum**:
The angular momentum can also be expressed in terms of the radius (\( r \)) and linear momentum (\( p \)):
\[ L = mvr \]
where \( m \) is the mass of the electron and \( v \) is its velocity. Setting these two expressions for angular momentum equal gives:
\[ mvr = n \frac{h}{2\pi} \]

4. **Solve for Velocity**:
Rearranging the equation for velocity, we have:
\[ v = \frac{n h}{2\pi mr} \]

5. **Substituting Momentum**:
The momentum \( p \) of the electron can be expressed as:
\[ p = mv \]
Substituting the expression for \( v \):
\[ p = m \left(\frac{n h}{2\pi mr}\right) = \frac{n h}{2\pi r} \]

6. **Substituting into the de-Broglie Wavelength Formula**:
Now substitute \( p \) back into the de-Broglie wavelength formula:
\[ \lambda = \frac{h}{p} = \frac{h}{\frac{n h}{2\pi r}} = \frac{2\pi r}{n} \]

7. **Calculate for the Ground State**:
For the ground state of hydrogen, \( n = 1 \):
\[ \lambda = 2\pi r \]

8. **Determine the Radius for the Ground State**:
The radius of the ground state (Bohr radius) is approximately:
\[ r = 5.29 \times 10^{-11} \text{ m} \]
Thus,
\[ \lambda = 2\pi (5.29 \times 10^{-11}) \approx 3.33 \times 10^{-10} \text{ m} \]

### Final Answer:
The de-Broglie wavelength of an electron in the ground state of the hydrogen atom is approximately \( 3.33 \times 10^{-10} \text{ m} \).