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(B) According to Heisenberg's uncertainty principle: $\Delta x \cdot \Delta p \geq \frac{h}{4\pi}$.Given: $\Delta p_x = \frac{1}{2} \Delta p_y$ and $\

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$2.5 	imes 10^{-10} \ 	ext{cm}$

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(B) According to Heisenberg's uncertainty principle: $\Delta x \cdot \Delta p \geq \frac{h}{4\pi}$.Given: $\Delta p_x = \frac{1}{2} \Delta p_y$ and $\Delta x_x = 0.05 \ \mathring{A}$.For particle $x$: $\Delta x_x \cdot \Delta p_x = \frac{h}{4\pi} \implies 0.05 \cdot \Delta p_x = \frac{h}{4\pi} \implies \Delta p_x = \frac{h}{4\pi \cdot 0.05}$.Since $\Delta p_x = \frac{1}{2} \Delta p_y$,then $\Delta p_y = 2 \cdot \Delta p_x = 2 \cdot \frac{h}{4\pi \cdot 0.05} = \frac{h}{4\pi \cdot 0.025}$.For particle $y$: $\Delta x_y \cdot \Delta p_y = \frac{h}{4\pi} \implies \Delta x_y = \frac{h}{4\pi \cdot \Delta p_y} = \frac{h}{4\pi \cdot (\frac{h}{4\pi \cdot 0.025})} = 0.025 \ \mathring{A}$.Converting to cm: $0.025 \ \mathring{A} = 0.025 	imes 10^{-8} \ 	ext{cm} = 2.5 	imes 10^{-10} \ 	ext{cm}$.

$x$ and $y$ are two moving particles. The uncertainty in the momentum of $x$ is half the uncertainty in the momentum of $y$. If the uncertainty in the position of particle $x$ is $0.05 \ \mathring{A}$,then the uncertainty in the position of particle $y$ is = .......

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