Problem

The speed of an electron is measured to be $5 \times 10^3 \, \mathrm{m/s}$ with an accuracy of $0.003\%$. Find the uncertainty in determining the position of this electron.

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Data

• Speed of electron: $v = 5 \times 10^3 \, \mathrm{m/s} \pm 0.003\%$
• Mass of electron: $m = 9.11 \times 10^{-31} \, \mathrm{kg}$
• Planck’s constant: $h = 6.626 \times 10^{-34} \, \mathrm{Js}$

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Prerequisite Concepts

1. Heisenberg’s Uncertainty Principle:

$$\Delta x \, \Delta p_x \geq \frac{h}{4\pi}$$

where $\Delta x$ is the uncertainty in position and $\Delta p_x$ is the uncertainty in momentum.

2. Momentum uncertainty:

$$\Delta p_x = m \Delta v$$

3. Speed uncertainty:

$$\Delta v = v \times \frac{\text{accuracy percentage}}{100}$$

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Solution

Step 1: Calculate the uncertainty in speed ($\Delta v$)

$$\Delta v = v \times \frac{0.003}{100}$$ $$\Delta v = 5 \times 10^3 \, \mathrm{m/s} \times \frac{0.003}{100}$$ $$\Delta v = 0.15 \, \mathrm{m/s}$$

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Step 2: Calculate the uncertainty in momentum ($\Delta p_x$)

Using $\Delta p_x = m \Delta v$:

$$\Delta p_x = (9.11 \times 10^{-31} \, \mathrm{kg}) \times (0.15 \, \mathrm{m/s})$$ $$\Delta p_x = 1.3665 \times 10^{-31} \, \mathrm{kg \cdot m/s}$$

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Step 3: Calculate the uncertainty in position ($\Delta x$)

Using Heisenberg’s uncertainty principle:

$$\Delta x = \frac{h}{\Delta p_x}$$

Substitute the values:

$$\Delta x = \frac{6.626 \times 10^{-34} \, \mathrm{Js}}{1.3665 \times 10^{-31} \, \mathrm{kg \cdot m/s}}$$ $$\Delta x = 4.85 \times 10^{-3} \, \mathrm{m}$$

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Answer

The uncertainty in the position of the electron is:

$$\Delta x = 4.85 \, \mathrm{mm}$$

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