A beam of specific kind of particles of velocity `2.1 xx 10^7 m//s` is scattered by a gold `(Z = 79)` nuclei. Find out specific charge (charge/mass) of this particle if the distance of closest approach is `2.5 xx 10^-14 m`.

Options

A. `4.84 xx 10^7 C//g`
B. `4.84 xx 10^-7 C//g`
C. `2.42 xx 10^7 C//g`
D. `3 xx 10^-12 C//g`

Explanation

To find the specific charge (charge/mass) of the particle that is scattered by the gold nuclei, we can follow these steps:

### Step 1: Understand the Concept of Closest Approach
At the distance of closest approach, the kinetic energy of the incoming particle is equal to the potential energy due to the electrostatic force between the particle and the gold nucleus.

### Step 2: Write the Equation for Kinetic Energy
The kinetic energy (KE) of the particle can be expressed as:
\[ KE = \frac{1}{2} mv^2 \]
where:
- \( m \) = mass of the particle
- \( v \) = velocity of the particle

Given \( v = 2.1 \times 10^7 \, \text{m/s} \).

### Step 3: Write the Equation for Potential Energy
The potential energy (PE) at the distance of closest approach can be expressed as:
\[ PE = \frac{k \cdot q_1 \cdot q_2}{r} \]
where:
- \( k \) = Coulomb's constant \( (9 \times 10^9 \, \text{N m}^2/\text{C}^2) \)
- \( q_1 \) = charge of the gold nucleus \( (Z \cdot e) \)
- \( q_2 \) = charge of the incoming particle
- \( r \) = distance of closest approach \( (2.5 \times 10^{-14} \, \text{m}) \)

For gold, \( Z = 79 \) and \( e = 1.6 \times 10^{-19} \, \text{C} \), thus:
\[ q_1 = Z \cdot e = 79 \cdot 1.6 \times 10^{-19} \, \text{C} \]

### Step 4: Set Kinetic Energy Equal to Potential Energy
At the closest approach:
\[ \frac{1}{2} mv^2 = \frac{k \cdot q_1 \cdot q_2}{r} \]

### Step 5: Solve for Specific Charge
Rearranging the equation to find the specific charge \( \frac{q_2}{m} \):
\[ \frac{q_2}{m} = \frac{v^2 \cdot r}{2k \cdot q_1} \]

### Step 6: Substitute the Known Values
Now we substitute the values into the equation:
- \( v = 2.1 \times 10^7 \, \text{m/s} \)
- \( r = 2.5 \times 10^{-14} \, \text{m} \)
- \( k = 9 \times 10^9 \, \text{N m}^2/\text{C}^2 \)
- \( q_1 = 79 \cdot 1.6 \times 10^{-19} \, \text{C} \)

Calculating \( q_1 \):
\[ q_1 = 79 \cdot 1.6 \times 10^{-19} = 1.264 \times 10^{-17} \, \text{C} \]

Now substituting everything into the specific charge equation:
\[ \frac{q_2}{m} = \frac{(2.1 \times 10^7)^2 \cdot (2.5 \times 10^{-14})}{2 \cdot (9 \times 10^9) \cdot (1.264 \times 10^{-17})} \]

### Step 7: Calculate the Result
Calculating the numerator:
\[ (2.1 \times 10^7)^2 = 4.41 \times 10^{14} \]
So,
\[ \text{Numerator} = 4.41 \times 10^{14} \cdot 2.5 \times 10^{-14} = 11.025 \times 10^{0} = 11.025 \]

Calculating the denominator:
\[ 2 \cdot (9 \times 10^9) \cdot (1.264 \times 10^{-17}) = 2.2768 \times 10^{-7} \]

Finally, calculating \( \frac{q_2}{m} \):
\[ \frac{q_2}{m} = \frac{11.025}{2.2768 \times 10^{-7}} \approx 4.84 \times 10^7 \, \text{C/kg} \]

### Conclusion
The specific charge of the particle is approximately:
\[ \frac{q_2}{m} \approx 4.84 \times 10^7 \, \text{C/kg} \]