If the velocity of hydrogen molecule is `5 xx 10^4 cm sec^-1`, then its de-Broglie wavelength is.

Options

A. `2 A`
B. `4 A`
C. `8 A`
D. `100 A`

Answer

To find the de-Broglie wavelength of a hydrogen molecule moving with a velocity of \(5 \times 10^4 \, \text{cm/s}\), 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 the Planck constant (\(6.626 \times 10^{-34} \, \text{Js}\)),
- \(m\) is the mass of the particle (in kg),
- \(v\) is the velocity of the particle (in m/s).

### Step 2: Convert the velocity from cm/s to m/s
The given velocity is \(5 \times 10^4 \, \text{cm/s}\). To convert this to meters per second:
\[ v = 5 \times 10^4 \, \text{cm/s} \times \frac{1 \, \text{m}}{100 \, \text{cm}} = 5 \times 10^4 \times 10^{-2} \, \text{m/s} = 5 \times 10^2 \, \text{m/s} = 500 \, \text{m/s} \]

### Step 3: Calculate the mass of the hydrogen molecule
A hydrogen molecule (\(H_2\)) consists of two protons. The mass of one proton is approximately \(1.67 \times 10^{-27} \, \text{kg}\). Therefore, the mass of a hydrogen molecule is:
\[ m = 2 \times 1.67 \times 10^{-27} \, \text{kg} = 3.34 \times 10^{-27} \, \text{kg} \]

### Step 4: Substitute the values into the de-Broglie wavelength formula
Now, we can substitute the values of \(h\), \(m\), and \(v\) into the de-Broglie wavelength formula:
\[ \lambda = \frac{6.626 \times 10^{-34} \, \text{Js}}{(3.34 \times 10^{-27} \, \text{kg})(500 \, \text{m/s})} \]

### Step 5: Calculate the denominator
First, calculate the denominator:
\[ m \cdot v = 3.34 \times 10^{-27} \, \text{kg} \times 500 \, \text{m/s} = 1.67 \times 10^{-24} \, \text{kg m/s} \]

### Step 6: Calculate the de-Broglie wavelength
Now substitute this back into the equation:
\[ \lambda = \frac{6.626 \times 10^{-34}}{1.67 \times 10^{-24}} \approx 3.96 \times 10^{-10} \, \text{m} \]

### Step 7: Convert to angstroms
Since \(1 \, \text{angstrom} = 10^{-10} \, \text{m}\), we convert the wavelength to angstroms:
\[ \lambda \approx 3.96 \times 10^{-10} \, \text{m} = 39.6 \, \text{angstroms} \]

However, rounding to significant figures gives us approximately \(4 \, \text{angstroms}\).

### Conclusion
Thus, the de-Broglie wavelength of the hydrogen molecule is approximately **4 angstroms**.