A tritrium gas target is bombared with a beam of monoenergetic protons of kinetic energy `K_(1) =3 MeV` The `KE` of the neutron emiited at `30^(@)` to the inicdent beam is `K_(2)` ? Find the value of `K_(1)//K_(2)` (approximately in whole number). Atomic masses are `H^(1) =1.007276 amu`, `n^(1)=1.008665 amu`, `._1 H^3=3.016050 amu`, `._2He^3=3.016030 amu`.

Explanation

To solve the problem, we need to analyze the nuclear reaction that occurs when a tritium gas target is bombarded with protons. The reaction can be represented as follows:

1. **Identify the Reaction**:
The reaction can be written as:
\[ ^3_1H + ^1_1p \rightarrow ^3_2He + n \]
where \( ^3_1H \) is tritium, \( ^1_1p \) is a proton, \( ^3_2He \) is helium-3, and \( n \) is a neutron.

2. **Calculate the Mass Defect**:
The mass defect (\( \Delta m \)) is calculated using the atomic masses provided:
\[ \Delta m = (m_{H} + m_{T}) - (m_{He} + m_{n}) \]
Substituting the values:
\[ m_{H} = 1.007276 \, \text{amu}, \quad m_{T} = 3.016050 \, \text{amu}, \quad m_{He} = 3.016030 \, \text{amu}, \quad m_{n} = 1.008665 \, \text{amu} \]
\[ \Delta m = (1.007276 + 3.016050) - (3.016030 + 1.008665) \]
\[ \Delta m = 4.023326 - 4.024695 = -0.001369 \, \text{amu} \]

3. **Convert Mass Defect to Energy**:
We convert the mass defect to energy using the conversion factor \( 1 \, \text{amu} = 931.5 \, \text{MeV} \):
\[ E = \Delta m \times 931.5 \, \text{MeV} \]
\[ E = -0.001369 \times 931.5 \approx -1.274 \, \text{MeV} \]
Since the energy is negative, it indicates that this is an endothermic reaction, meaning energy must be supplied for the reaction to occur.

4. **Calculate the Kinetic Energy of the Neutron**:
The kinetic energy of the emitted neutron (\( K_2 \)) can be calculated using conservation of momentum and energy. The neutron is emitted at an angle of \( 30^\circ \) to the incident beam. We can use the following formula:
\[ K_2 = \sqrt{u^2 + v^2} \]
where \( u \) and \( v \) are derived from the conservation laws.

5. **Calculate \( u \)**:
\[ u = \sqrt{\frac{m_H \cdot m_n \cdot K_1}{m_{He} + m_n}} \]
Substituting the values:
\[ K_1 = 3 \, \text{MeV}, \quad m_H = 1.007276 \, \text{amu}, \quad m_n = 1.008665 \, \text{amu}, \quad m_{He} = 3.016030 \, \text{amu} \]
\[ u = \sqrt{\frac{1.007276 \cdot 1.008665 \cdot 3}{3.016030 + 1.008665}} \cdot \cos(30^\circ) \]
\[ u \approx 0.375 \]

6. **Calculate \( v \)**:
\[ v = \sqrt{\frac{K_1 \cdot (m_{He} + m_H)}{m_{He} + m_n + m_H}} \]
Substituting the values:
\[ v \approx 0.542 \]

7. **Final Calculation of \( K_2 \)**:
\[ K_2 = \sqrt{u^2 + v^2} \approx \sqrt{(0.375)^2 + (0.542)^2} \approx 1.2 \, \text{MeV} \]

8. **Calculate the Ratio \( \frac{K_1}{K_2} \)**:
\[ \frac{K_1}{K_2} = \frac{3 \, \text{MeV}}{1.2 \, \text{MeV}} \approx 2.5 \]

Thus, the approximate value of \( \frac{K_1}{K_2} \) is **2.5**.