At `200^o` C, hydrogen molecules have velocity 5 ×` 10^4cm s^(−1)` . The de Broglie wavelength in this case is approximately :

Options

A. 1 `A^o`
B. 4 `A^o`
C. 6 `A^o`
D. 10 `A^o`

Answer

To find the de Broglie wavelength of hydrogen molecules 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:
\[ \lambda = \frac{h}{mv} \]
where:
- \( h \) is Planck's constant,
- \( m \) is the mass of the particle,
- \( v \) is the velocity of the particle.

### Step 2: Convert the given values
1. **Planck's constant (h)**:
- \( h = 6.626 \times 10^{-34} \, \text{Js} \) (or \( 6.626 \times 10^{-27} \, \text{ergs} \) if we are working in cgs units).

2. **Velocity (v)**:
- Given \( v = 5 \times 10^4 \, \text{cm/s} \).

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

Convert the mass to grams:
\[ m = \frac{2 \, \text{g/mol}}{6.022 \times 10^{23} \, \text{mol}^{-1}} = 3.32 \times 10^{-24} \, \text{g} \]
Convert to kilograms (since \( 1 \, \text{g} = 10^{-3} \, \text{kg} \)):
\[ m = 3.32 \times 10^{-24} \, \text{g} \times 10^{-3} \, \text{kg/g} = 3.32 \times 10^{-27} \, \text{kg} \]

### Step 4: Substitute values into the de Broglie wavelength formula
Now we can substitute the values into the de Broglie wavelength formula:
\[ \lambda = \frac{6.626 \times 10^{-34} \, \text{Js}}{(3.32 \times 10^{-27} \, \text{kg})(5 \times 10^4 \, \text{cm/s})} \]

### Step 5: Calculate the de Broglie wavelength
First, calculate the denominator:
\[ 3.32 \times 10^{-27} \, \text{kg} \times 5 \times 10^4 \, \text{cm/s} = 1.66 \times 10^{-22} \, \text{kg cm/s} \]

Now substitute this back into the wavelength equation:
\[ \lambda = \frac{6.626 \times 10^{-34}}{1.66 \times 10^{-22}} \approx 3.99 \times 10^{-12} \, \text{cm} \]

### Step 6: Convert to Angstroms
Since \( 1 \, \text{Å} = 10^{-8} \, \text{cm} \):
\[ \lambda \approx 3.99 \times 10^{-12} \, \text{cm} = 3.99 \times 10^{-4} \, \text{Å} \approx 4 \, \text{Å} \]

### Final Answer
The de Broglie wavelength of hydrogen molecules at 200°C is approximately:
\[ \lambda \approx 4 \, \text{Å} \]

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