At `200^o` C, hydrogen molecules have velocity 4 ×` 10^4cm s^(−1)` . The de Broglie wavelength in this case is approximately :

Options

A. 1 `A^o`
B. 6 `A^o`
C. 5 `A^o`
D. All of these

Explanation

To find the de Broglie wavelength of hydrogen molecules at 200°C with a given velocity, we will follow these steps:

### Step 1: Understand the Formula for de Broglie Wavelength
The de Broglie wavelength (λ) is given by the formula:
\[ \lambda = \frac{h}{mv} \]
where:
- \( h \) = Planck's constant
- \( m \) = mass of the particle
- \( v \) = velocity of the particle

### Step 2: Identify the Values
- The velocity \( v \) of hydrogen molecules is given as \( 4 \times 10^4 \) cm/s.
- Planck's constant \( h \) is approximately \( 6.6 \times 10^{-27} \) erg·s.
- The mass of one hydrogen molecule can be calculated using the formula:
\[ m = \frac{2}{N_a} \]
where \( N_a \) is Avogadro's number, approximately \( 6.02 \times 10^{23} \) molecules/mol.

### Step 3: Calculate the Mass of One Hydrogen Molecule
Substituting the values:
\[ m = \frac{2 \text{ g/mol}}{6.02 \times 10^{23} \text{ molecules/mol}} = \frac{2 \times 10^{-3} \text{ g}}{6.02 \times 10^{23}} = 3.32 \times 10^{-24} \text{ g} \]
To convert grams to grams per cm³ (since 1 g = \( 10^{-3} \) kg):
\[ m = 3.32 \times 10^{-24} \text{ g} = 3.32 \times 10^{-27} \text{ kg} \]

### Step 4: Substitute Values into the de Broglie Wavelength Formula
Now substituting \( h \), \( m \), and \( v \) into the de Broglie wavelength formula:
\[ \lambda = \frac{6.6 \times 10^{-27} \text{ erg·s}}{(3.32 \times 10^{-27} \text{ kg}) \times (4 \times 10^4 \text{ cm/s})} \]

### Step 5: Calculate the Wavelength
Calculating the denominator:
\[ m \cdot v = 3.32 \times 10^{-27} \times 4 \times 10^4 = 1.328 \times 10^{-22} \text{ kg·cm/s} \]
Now substituting back into the formula:
\[ \lambda = \frac{6.6 \times 10^{-27}}{1.328 \times 10^{-22}} \approx 4.96 \times 10^{-8} \text{ cm} \]

### Step 6: Convert to Angstroms
Since \( 1 \text{ Angstrom} = 10^{-8} \text{ cm} \):
\[ \lambda \approx 4.96 \text{ Angstroms} \approx 5 \text{ Angstroms} \]

### Final Answer
The de Broglie wavelength of hydrogen molecules at 200°C is approximately \( 5 \text{ Angstroms} \).

---