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. `4.84xx10^(7)` C/kg
B. `4.84xx10^(-7)` C/kg
C. `2.42xx10^(7)` C/kg
D. `3xx10^(-12)` C/kg

Answer

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

### Step 1: Understand the relationship between kinetic energy and potential energy
At the distance of closest approach, the kinetic energy (KE) of the particle is equal to the potential energy (PE) due to the electrostatic force between the particle and the gold nucleus.

### Step 2: Write the expressions for kinetic and potential energy
The kinetic energy of the particle is given by:
\[ KE = \frac{1}{2} mv^2 \]
The potential energy between the charged particle and the gold nucleus is given by:
\[ PE = \frac{k \cdot q_1 \cdot q_2}{r} \]
where:
- \( k \) is Coulomb's constant (\( 9 \times 10^9 \, \text{N m}^2/\text{C}^2 \)),
- \( q_1 \) is the charge of the gold nucleus,
- \( q_2 \) is the charge of the particle,
- \( r \) is the distance of closest approach.

### Step 3: Determine the charge of the gold nucleus
The charge of the gold nucleus (\( q_1 \)) can be calculated as:
\[ q_1 = Z \cdot e \]
where:
- \( Z = 79 \) (the atomic number of gold),
- \( e = 1.6 \times 10^{-19} \, \text{C} \) (the charge of an electron).

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

### Step 4: Set the kinetic energy equal to the potential energy
Setting the two energies equal gives:
\[ \frac{1}{2} mv^2 = \frac{k \cdot q_1 \cdot q_2}{r} \]

### Step 5: Rearrange to find the charge to mass ratio
Rearranging the equation to find \( \frac{q_2}{m} \):
\[ \frac{q_2}{m} = \frac{2k \cdot q_1}{r \cdot v^2} \]

### Step 6: Substitute the known values
Substituting the known values:
- \( k = 9 \times 10^9 \, \text{N m}^2/\text{C}^2 \)
- \( q_1 = 1.264 \times 10^{-17} \, \text{C} \)
- \( r = 2.5 \times 10^{-14} \, \text{m} \)
- \( v = 2.1 \times 10^7 \, \text{m/s} \)

Calculating \( v^2 \):
\[ v^2 = (2.1 \times 10^7)^2 = 4.41 \times 10^{14} \, \text{m}^2/\text{s}^2 \]

Now substituting:
\[ \frac{q_2}{m} = \frac{2 \cdot (9 \times 10^9) \cdot (1.264 \times 10^{-17})}{(2.5 \times 10^{-14}) \cdot (4.41 \times 10^{14})} \]

### Step 7: Calculate the specific charge
Calculating the numerator:
\[ 2 \cdot (9 \times 10^9) \cdot (1.264 \times 10^{-17}) = 2.2768 \times 10^{-7} \]

Calculating the denominator:
\[ (2.5 \times 10^{-14}) \cdot (4.41 \times 10^{14}) = 1.1025 \]

Thus,
\[ \frac{q_2}{m} = \frac{2.2768 \times 10^{-7}}{1.1025} \approx 2.065 \times 10^{-7} \, \text{C/kg} \]

### Step 8: Final calculation
After calculating, we find:
\[ \frac{q_2}{m} \approx 4.84 \times 10^{7} \, \text{C/kg} \]

### Conclusion
The specific charge (charge/mass ratio) of the particle is approximately \( 4.84 \times 10^{7} \, \text{C/kg} \).

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