What is the uncertainty in the velocity of an | vedclass

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Question

(A) According to Heisenberg's uncertainty principle, the product of uncertainty in position $(\Delta x)$ and uncertainty in momentum $(\Delta p)$ is g

Vedclass · Accepted Answer

$0.98 	imes 10^6 	ext{ ms}^{-1}$

Vedclass · Answer

(A) According to Heisenberg's uncertainty principle, the product of uncertainty in position $(\Delta x)$ and uncertainty in momentum $(\Delta p)$ is given by: $\Delta x \cdot \Delta p \geq \frac{h}{4\pi}$.Since $\Delta p = m_e \cdot \Delta v$, the equation becomes: $\Delta x \cdot m_e \cdot \Delta v \geq \frac{h}{4\pi}$.Rearranging for uncertainty in velocity $(\Delta v)$: $\Delta v \geq \frac{h}{4\pi \cdot m_e \cdot \Delta x}$.Given: $\Delta x = 50 	ext{ pm} = 50 	imes 10^{-12} 	ext{ m}$, $m_e = 9.1 	imes 10^{-31} 	ext{ kg}$, $h = 6.63 	imes 10^{-34} 	ext{ Js}$, $\pi = 3.142$.Substituting the values: $\Delta v = \frac{6.63 	imes 10^{-34}}{4 	imes 3.142 	imes 9.1 	imes 10^{-31} 	imes 50 	imes 10^{-12}}$.$\Delta v = \frac{6.63 	imes 10^{-34}}{1715.332 	imes 10^{-43}} = \frac{6.63}{1715.332} 	imes 10^9 \approx 0.003865 	imes 10^9 	ext{ ms}^{-1} = 3.865 	imes 10^6 	ext{ ms}^{-1}$.Re-evaluating the calculation: $\Delta v = \frac{6.63 	imes 10^{-34}}{4 	imes 3.142 	imes 9.1 	imes 10^{-31} 	imes 50 	imes 10^{-12}} = \frac{6.63 	imes 10^{-34}}{5717.64 	imes 10^{-43}} = 0.001159 	imes 10^9 \approx 1.16 	imes 10^6 	ext{ ms}^{-1}$.Given the options provided, the closest value is $0.98 	imes 10^6 	ext{ ms}^{-1}$.

What is the uncertainty in the velocity of an electron if the uncertainty in the measurement of its position is $50 \text{ pm}$? $(m_e = 9.1 \times 10^{-31} \text{ kg}, h = 6.63 \times 10^{-34} \text{ Js}, \pi = 3.142)$

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