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))`.

Explanation

To solve the problem, we will use the principle of conservation of energy. The total energy before the reaction must equal the total energy after the reaction. This includes the rest mass energy and the kinetic energy of the nuclei involved.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Kinetic energy of nucleus A, \( K_A = 5 \, \text{MeV} \)
- Kinetic energy of nucleus B, \( K_B = 3 \, \text{MeV} \)
- 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 Rest Mass Energies:**
- Rest mass energy of nucleus A:
\[ E_A = m_A \cdot 930 \, \text{MeV} = 25.0 \, \text{amu} \cdot 930 \, \text{MeV/amu} = 23250 \, \text{MeV} \]
- Rest mass energy of nucleus B:
\[ E_B = m_B \cdot 930 \, \text{MeV} = 10.0 \, \text{amu} \cdot 930 \, \text{MeV/amu} = 9300 \, \text{MeV} \]
- Rest mass energy of nucleus C:
\[ E_C = m_C \cdot 930 \, \text{MeV} = 34.995 \, \text{amu} \cdot 930 \, \text{MeV/amu} \approx 32500 \, \text{MeV} \]

3. **Apply Conservation of Energy:**
The total energy before the reaction is the sum of the rest mass energies and the kinetic energies of A and B:
\[ E_{\text{initial}} = E_A + E_B + K_A + K_B \]
Plugging in the values:
\[ E_{\text{initial}} = 23250 \, \text{MeV} + 9300 \, \text{MeV} + 5 \, \text{MeV} + 3 \, \text{MeV} = 32558 \, \text{MeV} \]

4. **Set Up the Energy Equation After the Reaction:**
After the reaction, the energy is given by the rest mass energy of C plus its kinetic energy and the excitation energy:
\[ E_{\text{final}} = E_C + K_C + E_{exc} \]
Rearranging gives:
\[ K_C = E_{\text{initial}} - E_C - E_{exc} \]

5. **Calculate the Kinetic Energy of Nucleus C:**
Substitute the known values into the equation:
\[ K_C = 32558 \, \text{MeV} - 32500 \, \text{MeV} - 10 \, \text{MeV} \]
\[ K_C = 32558 \, \text{MeV} - 32510 \, \text{MeV} = 48 \, \text{MeV} \]

6. **Final Result:**
The kinetic energy of nucleus C just after its formation is:
\[ K_C = 48 \, \text{MeV} \]