The de - Broglie wavelength associated with a hydrogen molecule moving with a thermal velocity of `3 km//s` will be

Options

A. `1 Å`
B. `0.66 Å`
C. `6.6 Å`
D. `66 Å`

Explanation

To find the de Broglie wavelength associated with a hydrogen molecule moving with a thermal velocity of \(3 \, \text{km/s}\), we can follow these steps:

### Step 1: Convert the velocity to meters per second
The given thermal velocity is \(3 \, \text{km/s}\). We need to convert this to meters per second (m/s).

\[ 3 \, \text{km/s} = 3 \times 10^3 \, \text{m/s} \]

### Step 2: Calculate the mass of a hydrogen molecule
A hydrogen molecule (\(H_2\)) consists of two hydrogen atoms. The mass of a single hydrogen atom is approximately \(1.67 \times 10^{-27} \, \text{kg}\). Therefore, the mass of a hydrogen molecule is:

\[ \text{Mass of } H_2 = 2 \times (1.67 \times 10^{-27} \, \text{kg}) = 3.34 \times 10^{-27} \, \text{kg} \]

### Step 3: Use the de Broglie wavelength formula
The de Broglie wavelength (\(\lambda\)) is given by the formula:

\[ \lambda = \frac{h}{p} \]

where \(h\) is Planck's constant and \(p\) is the momentum. The momentum \(p\) can be expressed as:

\[ p = mv \]

Substituting this into the de Broglie wavelength formula gives:

\[ \lambda = \frac{h}{mv} \]

### Step 4: Substitute the values into the formula
Planck's constant \(h\) is approximately \(6.626 \times 10^{-34} \, \text{Js}\). Now substituting the values:

\[ \lambda = \frac{6.626 \times 10^{-34} \, \text{Js}}{(3.34 \times 10^{-27} \, \text{kg})(3 \times 10^3 \, \text{m/s})} \]

### Step 5: Calculate the denominator
Calculating the denominator:

\[ 3.34 \times 10^{-27} \, \text{kg} \times 3 \times 10^3 \, \text{m/s} = 1.002 \times 10^{-23} \, \text{kg m/s} \]

### Step 6: Calculate the wavelength
Now substituting back into the wavelength formula:

\[ \lambda = \frac{6.626 \times 10^{-34}}{1.002 \times 10^{-23}} \approx 6.61 \times 10^{-11} \, \text{m} \]

### Step 7: Convert to Angstroms
Since \(1 \, \text{Angstrom} = 10^{-10} \, \text{m}\), we can convert the wavelength to Angstroms:

\[ \lambda \approx 0.661 \, \text{Angstroms} \]

### Final Answer
The de Broglie wavelength associated with a hydrogen molecule moving with a thermal velocity of \(3 \, \text{km/s}\) is approximately \(0.661 \, \text{Å}\).

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