At `200^o` C, hydrogen molecules have velocity 3.4 ×` 10^4cm s^(−1)` . The de Broglie wavelength in this case is approximately :

Options

A. 6 `A^o`
B. 1 `A^o`
C. 10 `A^o`
D. 1.8 `A^o`

Answer

To find the de Broglie wavelength of hydrogen molecules at 200°C with a velocity of \(3.4 \times 10^4 \, \text{cm/s}\), 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,
- \(m\) is the mass of the particle,
- \(v\) is the velocity of the particle.

### Step 2: Identify Planck's constant
Planck's constant (\(h\)) is approximately:
\[ h = 6.626 \times 10^{-34} \, \text{Js} \]
However, in the context of this problem, we will convert it to erg seconds for consistency with the mass unit we will use later:
\[ h = 6.626 \times 10^{-27} \, \text{erg s} \]

### Step 3: Calculate the mass of a hydrogen molecule
The mass of one molecule of hydrogen (\(H_2\)) can be calculated using:
\[ m = \frac{M}{N_a} \]
where:
- \(M\) is the molar mass of hydrogen (\(H_2\)), which is approximately \(2 \, \text{g/mol}\),
- \(N_a\) is Avogadro's number (\(6.022 \times 10^{23} \, \text{mol}^{-1}\)).

Converting the mass from grams to grams per erg (1 g = \(10^{7}\) erg):
\[ m = \frac{2 \times 10^{-3} \, \text{g}}{6.022 \times 10^{23}} = \frac{2 \times 10^{-3}}{6.022 \times 10^{23}} \, \text{g} \approx 3.32 \times 10^{-24} \, \text{g} \]

### Step 4: Convert mass to kg
To convert grams to kilograms (since \(1 \, \text{g} = 10^{-3} \, \text{kg}\)):
\[ m \approx 3.32 \times 10^{-24} \times 10^{-3} \, \text{kg} = 3.32 \times 10^{-27} \, \text{kg} \]

### Step 5: Substitute values into the de Broglie wavelength formula
Now we can substitute \(h\), \(m\), and \(v\) into the de Broglie wavelength formula:
\[ \lambda = \frac{6.626 \times 10^{-34} \, \text{Js}}{(3.32 \times 10^{-27} \, \text{kg})(3.4 \times 10^{4} \, \text{cm/s})} \]

### Step 6: Calculate the denominator
Calculating the denominator:
\[ (3.32 \times 10^{-27} \, \text{kg})(3.4 \times 10^{4} \, \text{cm/s}) \approx 1.13 \times 10^{-22} \, \text{kg cm/s} \]

### Step 7: Calculate the de Broglie wavelength
Now, substituting the values:
\[ \lambda \approx \frac{6.626 \times 10^{-34}}{1.13 \times 10^{-22}} \approx 5.86 \times 10^{-12} \, \text{cm} \]

### Step 8: Convert to angstroms
Since \(1 \, \text{Å} = 10^{-8} \, \text{cm}\):
\[ \lambda \approx 5.86 \times 10^{-12} \, \text{cm} = 5.86 \times 10^{-4} \, \text{Å} \approx 6 \, \text{Å} \]

### Final Result
Thus, the approximate de Broglie wavelength of hydrogen molecules at \(200^\circ C\) is:
\[ \lambda \approx 6 \, \text{Å} \]