A proton is accelerated to one tenth of the velocity of light. If its velocity can be measured with a precision `-pm1%`. What must be its uncertainty in position?

Explanation

To solve the problem, we will use Heisenberg's uncertainty principle, which states that the uncertainty in position (Δx) and the uncertainty in momentum (Δp) are related by the formula:

\[ \Delta x \cdot \Delta p \geq \frac{h}{4\pi} \]

where \( h \) is Planck's constant, approximately \( 6.626 \times 10^{-34} \, \text{J s} \).

### Step 1: Determine the velocity of the proton
The velocity of the proton is given as one-tenth of the speed of light (c). The speed of light is approximately \( 3 \times 10^8 \, \text{m/s} \). Therefore, the velocity (v) of the proton is:

\[ v = \frac{c}{10} = \frac{3 \times 10^8 \, \text{m/s}}{10} = 3 \times 10^7 \, \text{m/s} \]

### Step 2: Calculate the uncertainty in velocity (Δv)
The problem states that the velocity can be measured with a precision of ±1%. Therefore, the uncertainty in velocity (Δv) can be calculated as:

\[ \Delta v = 0.01 \times v = 0.01 \times (3 \times 10^7 \, \text{m/s}) = 3 \times 10^5 \, \text{m/s} \]

### Step 3: Calculate the momentum of the proton
The momentum (p) of the proton can be calculated using the formula:

\[ p = m \cdot v \]

The mass of a proton (m) is approximately \( 1.67 \times 10^{-27} \, \text{kg} \). Therefore, the momentum is:

\[ p = (1.67 \times 10^{-27} \, \text{kg}) \cdot (3 \times 10^7 \, \text{m/s}) = 5.01 \times 10^{-20} \, \text{kg m/s} \]

### Step 4: Calculate the uncertainty in momentum (Δp)
The uncertainty in momentum (Δp) can be calculated as:

\[ \Delta p = m \cdot \Delta v = (1.67 \times 10^{-27} \, \text{kg}) \cdot (3 \times 10^5 \, \text{m/s}) = 5.01 \times 10^{-22} \, \text{kg m/s} \]

### Step 5: Use Heisenberg's uncertainty principle to find Δx
Now, we can use the uncertainty principle to find the uncertainty in position (Δx):

\[ \Delta x \geq \frac{h}{4\pi \Delta p} \]

Substituting the values:

\[ \Delta x \geq \frac{6.626 \times 10^{-34} \, \text{J s}}{4\pi (5.01 \times 10^{-22} \, \text{kg m/s})} \]

Calculating the denominator:

\[ 4\pi (5.01 \times 10^{-22}) \approx 6.2832 \times 5.01 \times 10^{-22} \approx 3.15 \times 10^{-21} \]

Now substituting this value back into the equation for Δx:

\[ \Delta x \geq \frac{6.626 \times 10^{-34}}{3.15 \times 10^{-21}} \approx 2.10 \times 10^{-13} \, \text{m} \]

### Final Answer
Thus, the uncertainty in position (Δx) is approximately:

\[ \Delta x \approx 2.10 \times 10^{-13} \, \text{m} \]