If the velocity of hydrogen molecule is `5 xx 10^4 cm sec^-1`, then its de-Broglie wavelength is.

Options

A. `2 Å`
B. `4 Å`
C. `8 Å`
D. `100 Å`

Explanation

To find the de-Broglie wavelength of a hydrogen molecule (H₂) given its velocity, we can use the de-Broglie wavelength formula:

\[ \lambda = \frac{h}{mv} \]

where:
- \(\lambda\) is the de-Broglie wavelength,
- \(h\) is Planck's constant (\(6.626 \times 10^{-34} \, \text{Js}\)),
- \(m\) is the mass of the molecule,
- \(v\) is the velocity of the molecule.

### Step-by-Step Solution:

1. **Identify the given values**:
- Velocity of hydrogen molecule, \(v = 5 \times 10^4 \, \text{cm/s}\).
- Convert the velocity to meters per second:
\[ v = 5 \times 10^4 \, \text{cm/s} = 5 \times 10^4 \times \frac{1}{100} \, \text{m/s} = 5 \times 10^2 \, \text{m/s} = 500 \, \text{m/s} \]

2. **Calculate the mass of the hydrogen molecule (H₂)**:
- The mass of one hydrogen atom (proton) is approximately \(1.67 \times 10^{-27} \, \text{kg}\).
- Since H₂ consists of two hydrogen atoms, the mass of H₂ is:
\[ m = 2 \times 1.67 \times 10^{-27} \, \text{kg} = 3.34 \times 10^{-27} \, \text{kg} \]

3. **Substitute the values into the de-Broglie wavelength formula**:
\[ \lambda = \frac{6.626 \times 10^{-34} \, \text{Js}}{3.34 \times 10^{-27} \, \text{kg} \times 500 \, \text{m/s}} \]

4. **Calculate the denominator**:
\[ 3.34 \times 10^{-27} \, \text{kg} \times 500 \, \text{m/s} = 1.67 \times 10^{-24} \, \text{kg m/s} \]

5. **Calculate the de-Broglie wavelength**:
\[ \lambda = \frac{6.626 \times 10^{-34}}{1.67 \times 10^{-24}} \approx 3.96 \times 10^{-10} \, \text{m} \]

6. **Convert the wavelength to angstroms**:
- Since \(1 \, \text{angstrom} = 10^{-10} \, \text{m}\):
\[ \lambda \approx 3.96 \times 10^{-10} \, \text{m} = 3.96 \times 10^{1} \, \text{angstrom} \approx 4 \, \text{angstrom} \]

### Final Answer:
The de-Broglie wavelength of the hydrogen molecule is approximately \(4 \, \text{angstrom}\).