Question

Suppose that the momentum of a certain particle can be measured to an accuracy of one part in a thousand, Determine the minimum uncertainty in the position of a particle if the particle is (a) a 5 x 10-3 Kg mass moving with a speed of 2 m/s, (b) an electron moving with a speed of 1.8 x 108 m/s

Answer 1

Answer on Question # 68393-Physics / Quantum Mechanics

Suppose that the momentum of a certain particle can be measured to an accuracy of one part in a thousand. Determine the minimum uncertainty in the position of a particle if the particle is (a) a $5 \times 10^{-3} \mathrm{~kg}$ mass moving with a speed of $2 \mathrm{~m/s}$, (b) an electron moving with a speed of $1.8 \times 10^{-8} \mathrm{~m/s}$.

Solution

a) The momentum of the particle $p = m v = 5 \times 10^{-3} \times 2 = 1 \times 10^{-2} \, \mathrm{kg} \cdot \frac{\mathrm{m}}{\mathrm{s}}$.

So uncertainty in the momentum of a particle $\Delta p = \frac{1}{1000} p = 1 \times 10^{-5} \, \mathrm{kg} \cdot \frac{\mathrm{m}}{\mathrm{s}}$.

From the Heisenberg's uncertainty principle

$$\Delta p \Delta x \geq \frac{\hbar}{2}$$

the minimum uncertainty in the position of a particle

$$\Delta x = \frac{\hbar}{2 \Delta p} \approx \frac{1 \times 10^{-34}}{2 \times 10^{-5}} = 5 \times 10^{-28} \, \mathrm{m}.$$

b) The momentum of the relativistic electron

$$p = \frac{m v}{\sqrt{1 - \frac{v^{2}}{c^{2}}}} = \frac{9.1 \times 10^{-31} \times 1.8 \times 10^{8}}{\sqrt{1 - \frac{(1.8 \times 10^{8})^{2}}{(3 \times 10^{8})^{2}}}} = \frac{16.38 \times 10^{-23}}{0.8} \approx 2 \times 10^{-22} \, \mathrm{kg} \cdot \frac{\mathrm{m}}{\mathrm{s}}.$$ $$\Delta p = \frac{1}{1000} p = 2 \times 10^{-25} \, \mathrm{kg} \cdot \frac{\mathrm{m}}{\mathrm{s}}.$$

Finally

$$\Delta x = \frac{\hbar}{2 \Delta p} \approx \frac{1 \times 10^{-34}}{4 \times 10^{-25}} = 2.5 \times 10^{-10} \, \mathrm{m} = 2.5 \, \text{\AA}.$$

Answer a) $5 \times 10^{-28} \, \mathrm{m}$, b) $2.5 \times 10^{-10} \, \mathrm{m}$.

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