An `alpha`-particle of energy 4 Me V is scattered by gold foil (Z = 79) .Calculate the maximum volume in which positive charge of the atom is likely to be concentrated.

Options

A. `3.3 xx 10^(-40) m^(3)`
B. `7.7 xx 10^(-40) m^(3)`
C. `5.5 xx 10^(-40) m^(3)`
D. `11 xx 10^(-40) m^(3)`

Answer

To solve the problem of calculating the maximum volume in which the positive charge of the gold atom is likely to be concentrated when an alpha particle of energy 4 MeV is scattered, we will follow these steps:

### Step 1: Understand the Concept of Closest Approach
The closest approach \( r \) of an alpha particle to the nucleus can be derived from the energy conservation principle, where the kinetic energy of the alpha particle is converted into potential energy at the point of closest approach.

### Step 2: Formula for Closest Approach
The formula for the distance of closest approach \( r \) is given by:
\[ r = \frac{1}{4 \pi \epsilon_0} \cdot \frac{Z e^2}{E} \]
where:
- \( Z \) is the atomic number of the gold nucleus (79 for gold).
- \( e \) is the charge of an electron (\( 1.6 \times 10^{-19} \) C).
- \( E \) is the energy of the alpha particle in joules.

### Step 3: Convert Energy from MeV to Joules
The energy of the alpha particle is given as 4 MeV. We convert this to joules:
\[ E = 4 \, \text{MeV} = 4 \times 1.6 \times 10^{-13} \, \text{J} = 6.4 \times 10^{-13} \, \text{J} \]

### Step 4: Substitute Values into the Formula
Now, we substitute the values into the formula for \( r \):
\[ r = \frac{1}{4 \pi (9 \times 10^9)} \cdot \frac{79 \cdot (1.6 \times 10^{-19})^2}{6.4 \times 10^{-13}} \]

### Step 5: Calculate \( r \)
Calculating the numerator:
\[ 79 \cdot (1.6 \times 10^{-19})^2 = 79 \cdot 2.56 \times 10^{-38} = 2.0224 \times 10^{-36} \]
Now substituting into the equation:
\[ r = \frac{1}{4 \pi (9 \times 10^9)} \cdot \frac{2.0224 \times 10^{-36}}{6.4 \times 10^{-13}} \]
Calculating the denominator:
\[ 4 \pi (9 \times 10^9) \approx 113.097 \times 10^9 \approx 1.13097 \times 10^{11} \]
Thus:
\[ r = \frac{2.0224 \times 10^{-36}}{1.13097 \times 10^{11} \cdot 6.4 \times 10^{-13}} \approx 5.687 \times 10^{-24} \, \text{m} \]

### Step 6: Calculate the Volume
The volume \( V \) of the sphere in which the positive charge is concentrated is given by:
\[ V = \frac{4}{3} \pi r^3 \]
Substituting \( r \):
\[ V = \frac{4}{3} \pi (5.687 \times 10^{-24})^3 \]
Calculating \( r^3 \):
\[ (5.687 \times 10^{-24})^3 \approx 1.832 \times 10^{-70} \]
Thus:
\[ V \approx \frac{4}{3} \cdot \pi \cdot 1.832 \times 10^{-70} \approx 7.688 \times 10^{-70} \, \text{m}^3 \]

### Final Answer
The maximum volume in which the positive charge of the atom is likely to be concentrated is approximately:
\[ V \approx 7.688 \times 10^{-70} \, \text{m}^3 \]

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