Question

A proton is confined within a nucleus of size \( 10^{-13} \, \text{cm} \). The uncertainty in its velocity is ........... \( \times 10^8 \, \text{m/s} \).

Hint:

Use Heisenberg’s uncertainty principle to relate the uncertainty in position and momentum, and then solve for the uncertainty in velocity.

Answer 1

Correct Answer: 0.3

Step 1: Apply Heisenberg's uncertainty principle.
Heisenberg’s uncertainty principle gives a relation between the uncertainty in position \( \Delta x \) and the uncertainty in momentum \( \Delta p \): \[ \Delta x \Delta p \geq \frac{\hbar}{2} \] where \( \hbar = 1.055 \times 10^{-34} \, \text{J} \cdot \text{s} \). Since the proton is confined within a nucleus of size \( \Delta x = 10^{-13} \, \text{cm} = 10^{-15} \, \text{m} \), we can calculate the uncertainty in momentum \( \Delta p \).
Step 2: Calculate the uncertainty in momentum.
Rearranging the uncertainty principle to solve for \( \Delta p \): \[ \Delta p \geq \frac{\hbar}{2 \Delta x} \] Substitute the known values: \[ \Delta p \geq \frac{1.055 \times 10^{-34}}{2 \times 10^{-15}} = 5.275 \times 10^{-20} \, \text{kg} \cdot \text{m/s} \]
Step 3: Calculate the uncertainty in velocity.
The uncertainty in velocity \( \Delta v \) is related to the uncertainty in momentum \( \Delta p \) by: \[ \Delta p = m_p \Delta v \] where \( m_p = 1.672 \times 10^{-27} \, \text{kg} \) is the mass of the proton. Solving for \( \Delta v \): \[ \Delta v = \frac{\Delta p}{m_p} = \frac{5.275 \times 10^{-20}}{1.672 \times 10^{-27}} = 3.15 \times 10^7 \, \text{m/s} \]
Step 4: Conclusion.
Thus, the uncertainty in velocity is approximately 4.0 \( \times 10^8 \, \text{m/s} \).