Calculate the de-Broglie wavelength associated with an `alpha`-particle accelerated through a potential difference of 200 V. Given `m_(P) = 1.67 xx 10^(-27) kg`.

Answer

To calculate the de-Broglie wavelength associated with an alpha particle accelerated through a potential difference of 200 V, we can follow these steps:

### Step 1: Identify the known values
- Potential difference (V) = 200 V
- Mass of proton (m_P) = \(1.67 \times 10^{-27}\) kg
- Planck's constant (h) = \(6.626 \times 10^{-34}\) J·s
- Charge of an electron (e) = \(1.6 \times 10^{-19}\) C

### Step 2: Calculate the mass of the alpha particle
An alpha particle consists of 2 protons and 2 neutrons. Therefore, its mass can be calculated as:
\[ m_{\alpha} = 4 \times m_P = 4 \times (1.67 \times 10^{-27} \text{ kg}) = 6.68 \times 10^{-27} \text{ kg} \]

### Step 3: Calculate the charge of the alpha particle
The charge of an alpha particle is twice the charge of a proton (since it has 2 protons):
\[ q = 2 \times e = 2 \times (1.6 \times 10^{-19} \text{ C}) = 3.2 \times 10^{-19} \text{ C} \]

### Step 4: Use the de-Broglie wavelength formula
The de-Broglie wavelength (\(\lambda\)) can be calculated using the formula:
\[ \lambda = \frac{h}{\sqrt{2 m q V}} \]
Substituting the known values into the formula:
\[ \lambda = \frac{6.626 \times 10^{-34}}{\sqrt{2 \times (6.68 \times 10^{-27}) \times (3.2 \times 10^{-19}) \times (200)}} \]

### Step 5: Calculate the denominator
First, calculate the product inside the square root:
\[ 2 \times (6.68 \times 10^{-27}) \times (3.2 \times 10^{-19}) \times (200) = 2 \times 6.68 \times 3.2 \times 200 \times 10^{-46} \]
Calculating the numerical part:
\[ 2 \times 6.68 \times 3.2 \times 200 = 8,544 \text{ (approximately)} \]
Thus, the denominator becomes:
\[ \sqrt{8.544 \times 10^{-46}} \approx 2.93 \times 10^{-23} \]

### Step 6: Calculate the wavelength
Now substitute back into the wavelength equation:
\[ \lambda = \frac{6.626 \times 10^{-34}}{2.93 \times 10^{-23}} \approx 2.26 \times 10^{-11} \text{ m} \]

### Step 7: Convert to angstroms
Since \(1 \text{ angstrom} = 10^{-10} \text{ m}\):
\[ \lambda \approx 0.226 \text{ angstroms} \]

### Final Answer
The de-Broglie wavelength associated with the alpha particle accelerated through a potential difference of 200 V is approximately:
\[ \lambda \approx 0.226 \text{ Å} \]