At `200^(@)C`, hydrogen molecule have velocity `2.4 xx 10^(5)" cm s"^(-1)`. The de Broglie wavelength in this case is approximately.

Options

A. `1Å`
B. `1000Å`
C. `0.529xx(4)/(3)Å`
D. `10Å`

Explanation

To find the de Broglie wavelength of a hydrogen molecule at 200°C with a given velocity, 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 (\( 6.626 \times 10^{-34} \, \text{Js} \))
- \( p \) is the momentum of the particle.

### Step 2: Calculate the momentum (p)
The momentum \( p \) is calculated using the formula:
\[ p = mv \]
where:
- \( m \) is the mass of the hydrogen molecule.
- \( v \) is the velocity of the hydrogen molecule.

Given:
- Velocity \( v = 2.4 \times 10^5 \, \text{cm/s} = 2.4 \times 10^3 \, \text{m/s} \) (conversion from cm/s to m/s).

### Step 3: Find the mass of a hydrogen molecule
The mass of one molecule of hydrogen (H₂) can be calculated as follows:
\[ m = \frac{M}{N_A} \]
where:
- \( M \) is the molar mass of hydrogen (approximately \( 2 \, \text{g/mol} = 2 \times 10^{-3} \, \text{kg/mol} \)).
- \( N_A \) is Avogadro's number (\( 6.022 \times 10^{23} \, \text{mol}^{-1} \)).

Calculating the mass:
\[ m = \frac{2 \times 10^{-3} \, \text{kg/mol}}{6.022 \times 10^{23} \, \text{mol}^{-1}} \approx 3.32 \times 10^{-27} \, \text{kg} \]

### Step 4: Calculate the momentum (p)
Now substituting the values of \( m \) and \( v \) into the momentum formula:
\[ p = mv = (3.32 \times 10^{-27} \, \text{kg})(2.4 \times 10^3 \, \text{m/s}) \approx 7.97 \times 10^{-24} \, \text{kg m/s} \]

### Step 5: Calculate the de Broglie wavelength (λ)
Now substituting \( p \) into the de Broglie wavelength formula:
\[ \lambda = \frac{h}{p} = \frac{6.626 \times 10^{-34} \, \text{Js}}{7.97 \times 10^{-24} \, \text{kg m/s}} \approx 8.32 \times 10^{-11} \, \text{m} \]

### Step 6: Convert to Angstroms
To convert meters to Angstroms (1 Å = \( 10^{-10} \, \text{m} \)):
\[ \lambda \approx 8.32 \times 10^{-11} \, \text{m} = 0.832 \, \text{Å} \approx 0.8 \, \text{Å} \]

### Conclusion
Thus, the de Broglie wavelength of the hydrogen molecule at 200°C is approximately **1 Å**.

---