A beam of `alpha` particle moves with a velocity of `3.28 xx 10^(3) m s^(-1)` Calculate the wavelength of the `alpha` particles.

Explanation

To calculate the wavelength of an alpha particle moving with a given velocity, we can use the de Broglie wavelength formula:

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

where:
- \(\lambda\) is the wavelength,
- \(h\) is Planck's constant (\(6.63 \times 10^{-34} \, \text{Js}\)),
- \(m\) is the mass of the alpha particle,
- \(v\) is the velocity of the particle.

### Step 1: Identify the mass of the alpha particle
An alpha particle is essentially a helium nucleus, which consists of 2 protons and 2 neutrons. The approximate mass of an alpha particle is:

\[ m = 4 \times 1.66 \times 10^{-27} \, \text{kg} = 6.64 \times 10^{-27} \, \text{kg} \]

### Step 2: Identify the velocity of the alpha particle
The velocity of the alpha particle is given as:

\[ v = 3.28 \times 10^{3} \, \text{m/s} \]

### Step 3: Substitute the values into the de Broglie wavelength formula
Now we can substitute the values of \(h\), \(m\), and \(v\) into the formula:

\[ \lambda = \frac{6.63 \times 10^{-34} \, \text{Js}}{(6.64 \times 10^{-27} \, \text{kg}) \times (3.28 \times 10^{3} \, \text{m/s})} \]

### Step 4: Calculate the denominator
First, calculate the momentum \(mv\):

\[ mv = 6.64 \times 10^{-27} \, \text{kg} \times 3.28 \times 10^{3} \, \text{m/s} = 2.18 \times 10^{-23} \, \text{kg m/s} \]

### Step 5: Calculate the wavelength
Now substitute this value back into the wavelength formula:

\[ \lambda = \frac{6.63 \times 10^{-34} \, \text{Js}}{2.18 \times 10^{-23} \, \text{kg m/s}} \approx 3.04 \times 10^{-11} \, \text{m} \]

### Step 6: Convert to angstroms
Since \(1 \, \text{angstrom} = 10^{-10} \, \text{m}\):

\[ \lambda \approx 0.304 \, \text{angstroms} \]

### Final Answer
The wavelength of the alpha particles is approximately:

\[ \lambda \approx 0.304 \, \text{Å} \]