In the fission of `._(94)^(239) Pu` by a thermal neutron, two fission fragmnets of equal masses and sizes are produced and four neutrosn are emitted. Find the force between the two fission fragments at the moment they are produced.
Given : `R_(0) =1.`1 fermi.

Explanation

To find the force between the two fission fragments produced in the fission of \( _{94}^{239}Pu \) by a thermal neutron, we can follow these steps:

### Step 1: Determine the total mass number after fission
The total mass number before fission is the mass number of plutonium plus the mass number of the neutron:
\[ A_{\text{total}} = 239 + 1 = 240 \]

### Step 2: Calculate the mass number of each fission fragment
Since two fission fragments of equal mass are produced, we can divide the total mass number by 2:
\[ A = \frac{240}{2} = 120 \]

### Step 3: Determine the atomic number of each fission fragment
The total atomic number before fission is 94 (from plutonium). Since the fragments are equal, we can express the atomic number of each fragment as:
\[ 2Z = 94 \implies Z = \frac{94}{2} = 47 \]

### Step 4: Calculate the charge of each fission fragment
The charge \( Q \) of each fragment can be calculated using the formula:
\[ Q = Z \cdot e \]
where \( e \) is the elementary charge (\( e = 1.6 \times 10^{-19} \) coulombs):
\[ Q = 47 \cdot (1.6 \times 10^{-19}) = 7.52 \times 10^{-18} \text{ C} \]

### Step 5: Calculate the radius of each fission fragment
Using the formula for the radius of a nucleus:
\[ R = R_0 A^{1/3} \]
where \( R_0 = 1.1 \) fermi = \( 1.1 \times 10^{-15} \) m, we find:
\[ R = 1.1 \times 10^{-15} \cdot (120)^{1/3} \]
Calculating \( (120)^{1/3} \approx 4.93 \):
\[ R \approx 1.1 \times 10^{-15} \cdot 4.93 \approx 5.41 \times 10^{-15} \text{ m} \]

### Step 6: Calculate the distance between the centers of the two fragments
Since there are two fragments, the distance \( r \) between their centers is:
\[ r = 2R \approx 2 \cdot 5.41 \times 10^{-15} \approx 10.82 \times 10^{-15} \text{ m} \]

### Step 7: Calculate the force between the two fission fragments
Using Coulomb's law:
\[ F = \frac{k Q^2}{r^2} \]
where \( k = \frac{1}{4 \pi \epsilon_0} \approx 9 \times 10^9 \, \text{N m}^2/\text{C}^2 \):
\[ F = \frac{9 \times 10^9 \cdot (7.52 \times 10^{-18})^2}{(10.82 \times 10^{-15})^2} \]
Calculating the values:
\[ F = \frac{9 \times 10^9 \cdot 5.65 \times 10^{-35}}{1.17 \times 10^{-29}} \approx 4.35 \times 10^3 \text{ N} \]

### Final Answer
The force between the two fission fragments at the moment they are produced is approximately:
\[ F \approx 4.35 \times 10^3 \text{ N} \]