Question

The uncertainty in the velocities of two particles \( A \) and \( B \) are \( 0.03 \) and \( 0.01 \, \text{m/s} \), respectively. The mass of \( B \) is four times the mass of \( A \). The ratio of uncertainties in their positions is:

• A. $ \frac{4}{3} $
• B. $ \frac{3}{4} $
• C. $ \frac{16}{9} $
• D. $ \frac{9}{16} $

Hint:

When applying the Heisenberg Uncertainty Principle, remember that the uncertainty in position is inversely proportional to the product of mass and velocity uncertainty.

Answer 1

The Correct Option is A

Step 1: Known Information. 
Uncertainty in velocity of particle \( A \): \( \Delta v_A = 0.03 \, \text{m/s} \)
Uncertainty in velocity of particle \( B \): \( \Delta v_B = 0.01 \, \text{m/s} \)
Mass of particle \( B \) is four times the mass of particle \( A \): \[ m_B = 4m_A \] We need to find the ratio of uncertainties in their positions (\( \Delta x_A \) and \( \Delta x_B \)). 
Step 2: Heisenberg Uncertainty Principle. 
The Heisenberg Uncertainty Principle states: \[ \Delta x \cdot \Delta p \geq \frac{\hbar}{2} \] where:
\( \Delta x \) is the uncertainty in position,
\( \Delta p \) is the uncertainty in momentum,
\( \hbar = \frac{h}{2\pi} \) is the reduced Planck's constant.
Momentum is given by: \[ p = m v \] Thus, the uncertainty in momentum is: \[ \Delta p = m \Delta v \] Step 3: Apply the Uncertainty Principle for Each Particle. 
For particle \( A \):
\[ \Delta x_A \cdot \Delta p_A \geq \frac{\hbar}{2} \] Substitute \( \Delta p_A = m_A \Delta v_A \): \[ \Delta x_A \cdot m_A \Delta v_A \geq \frac{\hbar}{2} \] Rearrange to solve for \( \Delta x_A \): \[ \Delta x_A \geq \frac{\hbar}{2 m_A \Delta v_A} \] For particle \( B \): \[ \Delta x_B \cdot \Delta p_B \geq \frac{\hbar}{2} \] Substitute \( \Delta p_B = m_B \Delta v_B \): \[ \Delta x_B \cdot m_B \Delta v_B \geq \frac{\hbar}{2} \] Rearrange to solve for \( \Delta x_B \): \[ \Delta x_B \geq \frac{\hbar}{2 m_B \Delta v_B} \] Step 4: Ratio of Uncertainties in Position. 
The ratio of uncertainties in position is: \[ \frac{\Delta x_A}{\Delta x_B} = \frac{\frac{\hbar}{2 m_A \Delta v_A}}{\frac{\hbar}{2 m_B \Delta v_B}} \] Simplify: \[ \frac{\Delta x_A}{\Delta x_B} = \frac{\hbar}{2 m_A \Delta v_A} \cdot \frac{2 m_B \Delta v_B}{\hbar} \] \[ \frac{\Delta x_A}{\Delta x_B} = \frac{m_B \Delta v_B}{m_A \Delta v_A} \] Step 5: Substitute Given Values. 
\( m_B = 4m_A \)
\( \Delta v_A = 0.03 \, \text{m/s} \)
\( \Delta v_B = 0.01 \, \text{m/s} \)
Substitute these values into the ratio: \[ \frac{\Delta x_A}{\Delta x_B} = \frac{(4m_A)(0.01)}{m_A (0.03)} \] Simplify: \[ \frac{\Delta x_A}{\Delta x_B} = \frac{4 \cdot 0.01}{0.03} = \frac{0.04}{0.03} = \frac{4}{3} \] Final Answer: \( \boxed{\frac{4}{3}} \)