At `200^o` C, hydrogen molecules have velocity 1.4 ×` 10^5cm s^(−1)` . The de Broglie wavelength in this case is approximately :

Options

A. 10 `A^o`
B. 1 `A^o`
C. 100 `A^o`
D. 2 `A^o`

Explanation

To find the de Broglie wavelength of a hydrogen molecule at `200^o 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:
- \( h \) = Planck's constant \( = 6.626 \times 10^{-34} \, \text{Js} \)
- \( m \) = mass of the molecule
- \( v \) = velocity of the molecule

### Step 2: Convert the velocity to SI units
The given velocity is \( 1.4 \times 10^5 \, \text{cm/s} \). We need to convert this to meters per second (m/s):
\[ v = 1.4 \times 10^5 \, \text{cm/s} = 1.4 \times 10^5 \times \frac{1 \, \text{m}}{100 \, \text{cm}} = 1.4 \times 10^3 \, \text{m/s} \]

### 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:
- \( M \) = molar mass of hydrogen \( = 2 \, \text{g/mol} = 2 \times 10^{-3} \, \text{kg/mol} \)
- \( N_A \) = Avogadro's number \( = 6.022 \times 10^{23} \, \text{molecules/mol} \)

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

### Step 4: Substitute values into the de Broglie wavelength formula
Now we can substitute \( h \), \( m \), and \( v \) into the de Broglie wavelength formula:
\[ \lambda = \frac{6.626 \times 10^{-34} \, \text{Js}}{(3.32 \times 10^{-27} \, \text{kg})(1.4 \times 10^3 \, \text{m/s})} \]

### Step 5: Calculate the de Broglie wavelength
Calculating the denominator:
\[ mv = (3.32 \times 10^{-27} \, \text{kg})(1.4 \times 10^3 \, \text{m/s}) \approx 4.648 \times 10^{-24} \, \text{kg m/s} \]

Now substituting back:
\[ \lambda = \frac{6.626 \times 10^{-34}}{4.648 \times 10^{-24}} \approx 1.42 \times 10^{-10} \, \text{m} \]

### Step 6: Convert to centimeters
To convert meters to centimeters:
\[ \lambda \approx 1.42 \times 10^{-10} \, \text{m} = 1.42 \times 10^{-8} \, \text{cm} \]

### Step 7: Convert to Angstroms
Since \( 1 \, \text{Å} = 10^{-8} \, \text{cm} \):
\[ \lambda \approx 1.42 \, \text{Å} \]

### Conclusion
The de Broglie wavelength of the hydrogen molecule at `200^o C` is approximately \( 1.42 \, \text{Å} \).