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 (\( \lambda \)) 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 2: Use Bohr's quantization condition
According to Bohr's model of the hydrogen atom, the angular momentum of the electron in the nth orbit is quantized and given by:
\[ mvr = \frac{nh}{2\pi} \]
For the ground state of hydrogen, \( n = 1 \). Thus, we can rearrange this to find \( mv \):
\[ mv = \frac{h}{2\pi r} \]

### Step 3: Substitute \( mv \) into the de-Broglie wavelength formula
Substituting \( mv \) into the de-Broglie wavelength formula gives:
\[ \lambda = \frac{h}{mv} = \frac{h}{\frac{h}{2\pi r}} = 2\pi r \]

### Step 4: Calculate the de-Broglie wavelength for the first orbit
Given that the radius of the first orbit (\( r \)) is \( 0.53 \, \text{Å} \) (which is \( 0.53 \times 10^{-10} \, \text{m} \)), we can now calculate \( \lambda \):
\[ \lambda = 2\pi r = 2\pi \times 0.53 \times 10^{-10} \, \text{m} \]

### Step 5: Perform the calculation
Calculating this gives:
\[ \lambda = 2 \times 3.14 \times 0.53 \times 10^{-10} \approx 3.34 \times 10^{-10} \, \text{m} \]
Converting this to angstroms (1 Å = \( 10^{-10} \, \text{m} \)):
\[ \lambda \approx 3.34 \, \text{Å} \]

### Final Answer
The de-Broglie wavelength of the electron in the ground state of the hydrogen atom is approximately \( 3.34 \, \text{Å} \).

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