A beam of helium atoms moves with a velocity of `2xx10^(3) m s^(-1)`. Find the wavelength associated with helium atoms.

Explanation

To find the wavelength associated with helium atoms moving at a velocity of \(2 \times 10^3 \, \text{m/s}\), we can use the de Broglie wavelength formula:

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

where:
- \(\lambda\) is the wavelength,
- \(h\) is Planck's constant (\(6.626 \times 10^{-34} \, \text{J s}\)),
- \(m\) is the mass of the helium atom, and
- \(v\) is the velocity of the helium atoms.

### Step 1: Determine the mass of a helium atom

The molar mass of helium is approximately \(4 \, \text{g/mol}\). To find the mass of a single helium atom, we convert this to kilograms and divide by Avogadro's number (\(6.022 \times 10^{23} \, \text{atoms/mol}\)).

\[ \text{Mass of one helium atom} = \frac{4 \, \text{g/mol}}{6.022 \times 10^{23} \, \text{atoms/mol}} = \frac{4 \times 10^{-3} \, \text{kg}}{6.022 \times 10^{23}} \approx 6.64 \times 10^{-27} \, \text{kg} \]

### Step 2: Substitute values into the de Broglie wavelength formula

Now that we have the mass of a helium atom, we can substitute \(h\), \(m\), and \(v\) into the de Broglie equation.

\[ \lambda = \frac{6.626 \times 10^{-34} \, \text{J s}}{(6.64 \times 10^{-27} \, \text{kg})(2 \times 10^3 \, \text{m/s})} \]

### Step 3: Calculate the wavelength

Calculating the denominator:

\[ (6.64 \times 10^{-27} \, \text{kg})(2 \times 10^3 \, \text{m/s}) = 1.328 \times 10^{-23} \, \text{kg m/s} \]

Now substituting back into the equation for \(\lambda\):

\[ \lambda = \frac{6.626 \times 10^{-34}}{1.328 \times 10^{-23}} \approx 4.98 \times 10^{-11} \, \text{m} \]

### Final Answer

The wavelength associated with the helium atoms is approximately:

\[ \lambda \approx 4.98 \times 10^{-11} \, \text{m} \]

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