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 of determining the uncertainty in position of a proton accelerated to one-tenth of the velocity of light, we will use the Heisenberg Uncertainty Principle. Here’s a step-by-step solution:

### Step 1: Identify the Given Values
- The velocity of the proton, \( v = \frac{c}{10} = \frac{3 \times 10^8 \text{ m/s}}{10} = 3 \times 10^7 \text{ m/s} \)
- The precision in measuring the velocity, \( \Delta v = \pm 1\% \) of \( v \)

### Step 2: Calculate the Uncertainty in Velocity
To find the uncertainty in velocity, we calculate:
\[ \Delta v = 0.01 \times v = 0.01 \times 3 \times 10^7 \text{ m/s} = 3 \times 10^5 \text{ m/s} \]

### Step 3: Use the Heisenberg Uncertainty Principle
The Heisenberg Uncertainty Principle states:
\[ \Delta x \cdot \Delta p \geq \frac{h}{4\pi} \]
Where:
- \( \Delta x \) = uncertainty in position
- \( \Delta p \) = uncertainty in momentum
- \( h \) = Planck's constant \( \approx 6.63 \times 10^{-34} \text{ J s} \)

### Step 4: Relate Uncertainty in Momentum to Uncertainty in Velocity
Momentum \( p \) is given by:
\[ p = m \cdot v \]
Thus, the uncertainty in momentum \( \Delta p \) can be expressed as:
\[ \Delta p = m \cdot \Delta v \]
Where \( m \) is the mass of the proton, approximately \( 1.67 \times 10^{-27} \text{ kg} \).

### Step 5: Substitute Values into the Uncertainty Principle
Substituting \( \Delta p \) into the uncertainty principle:
\[ \Delta x \cdot (m \cdot \Delta v) \geq \frac{h}{4\pi} \]
Rearranging gives:
\[ \Delta x \geq \frac{h}{4\pi m \Delta v} \]

### Step 6: Plug in the Values
Now we can substitute the values:
- \( h = 6.63 \times 10^{-34} \text{ J s} \)
- \( m = 1.67 \times 10^{-27} \text{ kg} \)
- \( \Delta v = 3 \times 10^5 \text{ m/s} \)

Calculating:
\[ \Delta x \geq \frac{6.63 \times 10^{-34}}{4 \times 3.14 \times 1.67 \times 10^{-27} \times 3 \times 10^5} \]

### Step 7: Calculate the Result
Calculating the denominator:
\[ 4 \times 3.14 \times 1.67 \times 10^{-27} \times 3 \times 10^5 \approx 3.14 \times 10^{-21} \]
Now substituting back:
\[ \Delta x \geq \frac{6.63 \times 10^{-34}}{3.14 \times 10^{-21}} \approx 2.11 \times 10^{-13} \text{ m} \]

### Final Result
Thus, the uncertainty in position \( \Delta x \) is approximately:
\[ \Delta x \approx 2.11 \times 10^{-13} \text{ m} \]