A uranium 238 nucleus, initially at rest emits an alpha particle with a speed of `1.4xx10^7m/s`. Calculate the recoil speed of the residual nucleus thorium 234. Assume that the mas of a nucleus is proportional to the mass number.

Explanation

To solve the problem of calculating the recoil speed of the residual nucleus (thorium-234) after a uranium-238 nucleus emits an alpha particle, we will apply the principle of conservation of momentum.

### Step-by-Step Solution:

1. **Identify the Given Information:**
- Mass of uranium-238 nucleus (U-238) = 238 units (proportional to mass number)
- Mass of alpha particle (He-4) = 4 units (proportional to mass number)
- Speed of the emitted alpha particle (v_alpha) = \(1.4 \times 10^7 \, \text{m/s}\)

2. **Determine the Mass of the Residual Nucleus:**
- Mass of the residual nucleus (thorium-234) = Mass of U-238 - Mass of alpha particle
- Mass of Th-234 = \(238 - 4 = 234\) units

3. **Apply Conservation of Momentum:**
- Initial momentum of the system (U-238 at rest) = 0
- Final momentum of the system = Momentum of alpha particle + Momentum of thorium-234
- Mathematically, this can be expressed as:
\[ 0 = (m_{\alpha} \cdot v_{\alpha}) + (m_{Th} \cdot v_{Th}) \]
- Where:
- \(m_{\alpha} = 4\) (mass of alpha particle)
- \(v_{\alpha} = 1.4 \times 10^7 \, \text{m/s}\)
- \(m_{Th} = 234\) (mass of thorium-234)
- \(v_{Th}\) = recoil speed of thorium-234 (to be calculated)

4. **Set Up the Equation:**
- Plugging in the values:
\[ 0 = (4 \cdot 1.4 \times 10^7) + (234 \cdot v_{Th}) \]
- Rearranging gives:
\[ 234 \cdot v_{Th} = - (4 \cdot 1.4 \times 10^7) \]

5. **Calculate the Recoil Speed:**
- Now, calculate \(v_{Th}\):
\[ v_{Th} = - \frac{4 \cdot 1.4 \times 10^7}{234} \]
- Performing the calculation:
\[ v_{Th} = - \frac{5.6 \times 10^7}{234} \approx -2.39 \times 10^5 \, \text{m/s} \]

6. **Final Result:**
- The negative sign indicates that the direction of the recoil speed is opposite to that of the emitted alpha particle.
- Thus, the recoil speed of the thorium-234 nucleus is approximately \(2.39 \times 10^5 \, \text{m/s}\).