The uncertainty in the velocities of two part | vedclass

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(A) According to Heisenberg's uncertainty principle,$\Delta x \cdot \Delta p \ge \frac{h}{4\pi}$.Since $\Delta p = m \cdot \Delta v$,we have $\Delta x

Vedclass · Accepted Answer

$4/3$

Vedclass · Answer

(A) According to Heisenberg's uncertainty principle,$\Delta x \cdot \Delta p \ge \frac{h}{4\pi}$.Since $\Delta p = m \cdot \Delta v$,we have $\Delta x \cdot m \cdot \Delta v = 	ext{constant}$.Therefore,$\Delta x \propto \frac{1}{m \cdot \Delta v}$.Given $m_B = 4m_A$,$\Delta v_A = 0.03 \ m \ s^{-1}$,and $\Delta v_B = 0.01 \ m \ s^{-1}$.The ratio of uncertainties in positions is $\frac{\Delta x_A}{\Delta x_B} = \frac{m_B \cdot \Delta v_B}{m_A \cdot \Delta v_A}$.Substituting the values: $\frac{\Delta x_A}{\Delta x_B} = \frac{4m_A \cdot 0.01}{m_A \cdot 0.03} = \frac{4 \cdot 0.01}{0.03} = \frac{4}{3}$.

The uncertainty in the velocities of two particles $A$ and $B$ are $0.03 \ m \ s^{-1}$ and $0.01 \ m \ s^{-1}$ respectively. The mass of $B$ is four times the mass of $A$. The ratio of uncertainties in their positions is

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