Consider a nuclear reaction `A+B rarr C`. A nucleus `A` moving with kinetic energy of `5 MeV` collides with a nucleus `B` moving with kinetic energy of `3MeV` and forms a nucleus `C` in exicted state. Find the kinetic energy of nucleus `C` just after its fromation if it is formed in a state with excitation energy `10 MeV`. Take masses of nuclei of `A,B` and `C ` as `25.0,10.0,34.995 amu`, respectively.
(1amu=`930 MeV//c^(2))`.

Answer

To solve the problem, we will use the conservation of energy principle. The total energy before the reaction must equal the total energy after the reaction.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Kinetic energy of nucleus A, \( KE_A = 5 \, \text{MeV} \)
- Kinetic energy of nucleus B, \( KE_B = 3 \, \text{MeV} \)
- Excitation energy of nucleus C, \( E_{exc} = 10 \, \text{MeV} \)
- Mass of nucleus A, \( m_A = 25.0 \, \text{amu} \)
- Mass of nucleus B, \( m_B = 10.0 \, \text{amu} \)
- Mass of nucleus C, \( m_C = 34.995 \, \text{amu} \)
- Conversion factor: \( 1 \, \text{amu} = 930 \, \text{MeV}/c^2 \)

2. **Calculate the Total Initial Energy:**
The total initial energy is the sum of the rest mass energy and the kinetic energies of both nuclei:
\[ E_{initial} = (m_A + m_B) \times 930 \, \text{MeV} + KE_A + KE_B \]
Substituting the values:
\[ E_{initial} = (25.0 + 10.0) \times 930 + 5 + 3 \]
\[ E_{initial} = 35.0 \times 930 + 8 = 32550 + 8 = 32558 \, \text{MeV} \]

3. **Calculate the Total Final Energy:**
The total final energy consists of the rest mass energy of nucleus C and its kinetic energy:
\[ E_{final} = m_C \times 930 + KE_C + E_{exc} \]
Rearranging gives us:
\[ KE_C = E_{final} - (m_C \times 930 + E_{exc}) \]

4. **Calculate the Rest Mass Energy of Nucleus C:**
\[ m_C \times 930 = 34.995 \times 930 = 32525.35 \, \text{MeV} \]

5. **Set Up the Equation for Conservation of Energy:**
\[ E_{initial} = E_{final} \]
\[ 32558 = 32525.35 + KE_C + 10 \]
Rearranging gives:
\[ KE_C = 32558 - 32525.35 - 10 \]
\[ KE_C = 32558 - 32535.35 = 22.65 \, \text{MeV} \]

6. **Final Calculation:**
\[ KE_C = 22.65 \, \text{MeV} \]

### Final Answer:
The kinetic energy of nucleus C just after its formation is \( KE_C = 22.65 \, \text{MeV} \).