If the radius of the first orbit of the hydrogen atom is `0.53 Å` , then the de-Broglie wavelength of the electron in the ground state of hydrogen atom will be

Options

A. `0.53 Å`
B. `3.33 Å`
C. `1.67 Å`
D. `1.06 Å`

Answer

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

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

### Step 2: Relate momentum to mass and velocity
Momentum \( p \) can be expressed as:
\[ p = mv \]
where \( m \) is the mass of the electron and \( v \) is its velocity.

### Step 3: Use Bohr's model to find the velocity
According to Bohr's model, the angular momentum of the electron in the nth orbit is quantized:
\[ mvr = \frac{n h}{2 \pi} \]
For the ground state of hydrogen, \( n = 1 \) and the radius \( r \) is given as \( 0.53 \, \text{Å} \) (or \( 0.53 \times 10^{-10} \, \text{m} \)).

### Step 4: Solve for velocity
From the angular momentum equation, we can express the velocity \( v \) as:
\[ v = \frac{n h}{2 \pi m r} \]
Substituting \( n = 1 \):
\[ v = \frac{h}{2 \pi m r} \]

### Step 5: Substitute \( v \) into the momentum equation
Now substituting \( v \) back into the momentum equation:
\[ p = mv = m \left( \frac{h}{2 \pi m r} \right) \]
This simplifies to:
\[ p = \frac{h}{2 \pi r} \]

### Step 6: Substitute momentum into the de-Broglie wavelength formula
Now substituting \( p \) into the de-Broglie wavelength formula:
\[ \lambda = \frac{h}{p} = \frac{h}{\frac{h}{2 \pi r}} = 2 \pi r \]

### Step 7: Substitute the radius value
Now, we can substitute the radius \( r = 0.53 \, \text{Å} = 0.53 \times 10^{-10} \, \text{m} \):
\[ \lambda = 2 \pi (0.53 \times 10^{-10}) \]
Calculating this gives:
\[ \lambda = 2 \times 3.14 \times 0.53 \times 10^{-10} \]
\[ \lambda \approx 3.34 \times 10^{-10} \, \text{m} = 3.34 \, \text{Å} \]

### Final Answer
Thus, the de-Broglie wavelength of the electron in the ground state of the hydrogen atom is approximately:
\[ \lambda \approx 3.34 \, \text{Å} \]