At `200^o` C, hydrogen molecules have velocity 7 ×` 10^4cm s^(−1)` . The de Broglie wavelength in this case is approximately :

Options

A. 3 `A^o`
B. 1 `A^o`
C. 4 `A^o`
D. 5 `A^o`

Explanation

To find the de Broglie wavelength of hydrogen molecules at 200°C with a velocity of \(7 \times 10^4 \, \text{cm/s}\), 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 molecule,
- \(v\) is the velocity of the molecule.

### Step 2: Convert the mass of a hydrogen molecule
The mass of a hydrogen molecule (H₂) can be calculated using Avogadro's number (\(N_a = 6.022 \times 10^{23} \, \text{mol}^{-1}\)). The molar mass of hydrogen (H₂) is approximately \(2 \, \text{g/mol}\), so:
\[ m = \frac{2 \, \text{g/mol}}{N_a} = \frac{2 \times 10^{-3} \, \text{kg/mol}}{6.022 \times 10^{23} \, \text{mol}^{-1}} \approx 3.32 \times 10^{-27} \, \text{kg} \]

### Step 3: Convert the velocity to SI units
The given velocity is \(7 \times 10^4 \, \text{cm/s}\). To convert this to meters per second:
\[ v = 7 \times 10^4 \, \text{cm/s} = 7 \times 10^4 \times 10^{-2} \, \text{m/s} = 7 \times 10^2 \, \text{m/s} = 700 \, \text{m/s} \]

### Step 4: Substitute values into the de Broglie wavelength formula
Now we can substitute the values of \(h\), \(m\), and \(v\) into the de Broglie wavelength formula:
\[ \lambda = \frac{6.626 \times 10^{-34} \, \text{Js}}{3.32 \times 10^{-27} \, \text{kg} \times 700 \, \text{m/s}} \]

### Step 5: Calculate the denominator
Calculating the denominator:
\[ 3.32 \times 10^{-27} \, \text{kg} \times 700 \, \text{m/s} = 2.324 \times 10^{-24} \, \text{kg m/s} \]

### Step 6: Calculate the de Broglie wavelength
Now substituting back into the equation:
\[ \lambda = \frac{6.626 \times 10^{-34}}{2.324 \times 10^{-24}} \approx 2.85 \times 10^{-10} \, \text{m} \]

### Step 7: Convert to centimeters
To convert meters to centimeters:
\[ \lambda \approx 2.85 \times 10^{-10} \, \text{m} \times 100 \, \text{cm/m} = 2.85 \times 10^{-8} \, \text{cm} \]

### Step 8: Express in angstroms
Since \(1 \, \text{Å} = 10^{-8} \, \text{cm}\):
\[ \lambda \approx 2.85 \, \text{Å} \approx 3 \, \text{Å} \]

### Final Answer
Thus, the de Broglie wavelength of hydrogen molecules at \(200^\circ C\) is approximately:
\[ \lambda \approx 3 \, \text{Å} \]

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