A helium atom at 300 K is moving with a velocity of `2.40 xx 10^(2) ms^(-1)`. The de-Broglie wavelength is about [At. Wt. of He=4.0]

Options

A. 0.416 nm
B. 0.83 nm
C. 803 Å
D. `8000 Å`

Answer

To find the de-Broglie wavelength of a helium atom moving with a given velocity, we can use the de-Broglie wavelength formula:

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

where:
- \(\lambda\) is the de-Broglie wavelength,
- \(h\) is the Planck’s constant (\(6.626 \times 10^{-34} \, \text{Js}\)),
- \(m\) is the mass of the helium atom in kilograms,
- \(v\) is the velocity of the atom.

### Step 1: Find the mass of a helium atom
The atomic weight of helium is given as 4.0 g/mol. To find the mass of a single helium atom, we convert this to kilograms and divide by Avogadro's number (\(N_A = 6.022 \times 10^{23} \, \text{mol}^{-1}\)).

\[ m = \frac{4.0 \, \text{g/mol}}{N_A} = \frac{4.0 \times 10^{-3} \, \text{kg/mol}}{6.022 \times 10^{23} \, \text{mol}^{-1}} \approx 6.64 \times 10^{-27} \, \text{kg} \]

### Step 2: Substitute values into the de-Broglie wavelength formula
Now we can substitute the values into the de-Broglie wavelength formula. The velocity \(v\) is given as \(2.40 \times 10^{2} \, \text{m/s}\).

\[ \lambda = \frac{6.626 \times 10^{-34} \, \text{Js}}{(6.64 \times 10^{-27} \, \text{kg})(2.40 \times 10^{2} \, \text{m/s})} \]

### Step 3: Calculate the denominator
Calculating the denominator:

\[ m \cdot v = 6.64 \times 10^{-27} \, \text{kg} \times 2.40 \times 10^{2} \, \text{m/s} \approx 1.5936 \times 10^{-24} \, \text{kg m/s} \]

### Step 4: Calculate the de-Broglie wavelength
Now substituting back into the formula:

\[ \lambda = \frac{6.626 \times 10^{-34}}{1.5936 \times 10^{-24}} \approx 4.15 \times 10^{-10} \, \text{m} \]

### Step 5: Convert to nanometers
To convert meters to nanometers, we multiply by \(10^9\):

\[ \lambda \approx 4.15 \times 10^{-10} \, \text{m} \times 10^9 \, \text{nm/m} \approx 0.415 \, \text{nm} \]

### Conclusion
The de-Broglie wavelength of the helium atom is approximately \(0.415 \, \text{nm}\).