To find Chadwick's value for the neutron mass using the provided equation and the given data, we will follow these steps:
### Step 1: Identify the Variables
We need to identify the variables based on the problem statement:
- For the first target (hydrogen):
- Mass \( m_2 = 1u \) (mass of proton)
- Maximum recoil speed \( v_2 = 3.3 \times 10^7 \, \text{m/s} \)
- For the second target (nitrogen):
- Mass \( m_1 = 14u \) (mass of nitrogen nucleus)
- Maximum recoil speed \( v_1 = 4.7 \times 10^6 \, \text{m/s} \)
### Step 2: Write Down the Formula
The formula for calculating the neutron mass \( m_n \) is given by:
\[
m_n = \frac{m_1 v_1 - m_2 v_2}{v_2 - v_1}
\]
### Step 3: Substitute the Values into the Formula
Substituting the known values into the equation:
\[
m_n = \frac{(14u)(4.7 \times 10^6 \, \text{m/s}) - (1u)(3.3 \times 10^7 \, \text{m/s})}{(3.3 \times 10^7 \, \text{m/s}) - (4.7 \times 10^6 \, \text{m/s})}
\]
### Step 4: Calculate the Numerator
Calculating the numerator:
\[
14u \cdot 4.7 \times 10^6 \, \text{m/s} = 65.8u \times 10^6 \, \text{m/s}
\]
\[
1u \cdot 3.3 \times 10^7 \, \text{m/s} = 33.0u \times 10^6 \, \text{m/s}
\]
Thus,
\[
\text{Numerator} = 65.8u \times 10^6 \, \text{m/s} - 33.0u \times 10^6 \, \text{m/s} = 32.8u \times 10^6 \, \text{m/s}
\]
### Step 5: Calculate the Denominator
Calculating the denominator:
\[
(3.3 \times 10^7 \, \text{m/s}) - (4.7 \times 10^6 \, \text{m/s}) = 33.0 \times 10^6 \, \text{m/s} - 4.7 \times 10^6 \, \text{m/s} = 28.3 \times 10^6 \, \text{m/s}
\]
### Step 6: Combine the Results
Now substituting back into the equation:
\[
m_n = \frac{32.8u \times 10^6 \, \text{m/s}}{28.3 \times 10^6 \, \text{m/s}}
\]
### Step 7: Simplify the Expression
The \( 10^6 \, \text{m/s} \) cancels out:
\[
m_n = \frac{32.8u}{28.3} \approx 1.16u
\]
### Conclusion
Chadwick's value for the neutron mass is approximately:
\[
m_n \approx 1.16u
\]
