To solve the problem, we will use the principles of quantum mechanics, specifically the Heisenberg uncertainty principle, which states that the product of the uncertainties in position (Δx) and momentum (Δp) is at least on the order of Planck's constant divided by 2:
\[
\Delta x \cdot \Delta p \geq \frac{h}{2}
\]
Where:
- \( \Delta x \) is the uncertainty in position,
- \( \Delta p \) is the uncertainty in momentum,
- \( h \) is Planck's constant, approximately \( 6.626 \times 10^{-34} \, \text{J s} \).
### Step-by-Step Solution:
1. **Calculate the momentum (p₀) of the proton**:
The momentum \( p \) is given by the formula:
\[
p = mv
\]
Where:
- \( m \) is the mass of the proton, approximately \( 1.673 \times 10^{-27} \, \text{kg} \),
- \( v \) is the speed of the proton, given as \( 0.45 \, \text{Mms}^{-1} = 0.45 \times 10^{6} \, \text{m/s} = 4.5 \times 10^{5} \, \text{m/s} \).
So,
\[
p_0 = (1.673 \times 10^{-27} \, \text{kg}) \times (4.5 \times 10^{5} \, \text{m/s}) = 7.5135 \times 10^{-22} \, \text{kg m/s}
\]
2. **Calculate the uncertainty in momentum (Δp)**:
The problem states that the uncertainty in momentum is to be reduced to \( 0.0100\% \) of \( p_0 \):
\[
\Delta p = 0.0100\% \times p_0 = 0.0001 \times p_0 = 0.0001 \times 7.5135 \times 10^{-22} \, \text{kg m/s}
\]
\[
\Delta p = 7.5135 \times 10^{-26} \, \text{kg m/s}
\]
3. **Use the Heisenberg uncertainty principle to find Δx**:
Rearranging the uncertainty principle gives us:
\[
\Delta x \geq \frac{h}{2 \Delta p}
\]
Substituting the values:
\[
\Delta x \geq \frac{6.626 \times 10^{-34} \, \text{J s}}{2 \times 7.5135 \times 10^{-26} \, \text{kg m/s}}
\]
\[
\Delta x \geq \frac{6.626 \times 10^{-34}}{1.5027 \times 10^{-25}} \approx 4.41 \times 10^{-9} \, \text{m}
\]
4. **Convert Δx to nanometers**:
To convert meters to nanometers:
\[
\Delta x \approx 4.41 \times 10^{-9} \, \text{m} = 0.441 \, \text{nm}
\]
### Final Answer:
The uncertainty in the location that must be tolerated is approximately \( 0.441 \, \text{nm} \).
