To find the uncertainty in the position of the proton, we use the Heisenberg Uncertainty Principle, which in terms of position ($$x$$) and momentum ($$p$$) is expressed as:

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

Where $$\hbar$$ (the reduced Planck constant) is approximately $$1.05 \times 10^{-34} \mathrm{~Js}$$.

First, let's calculate the momentum of the proton when it's moving at one tenth of the velocity of light ($$c = 3.00 \times 10^8 \mathrm{~m/s}$$).

The velocity of the proton, $$v = 0.1c = 0.1 \times 3.00 \times 10^8 \mathrm{~m/s} = 3.00 \times 10^7 \mathrm{~m/s}$$.

The momentum $$p$$ of the proton is given by:

$$p = mv$$

where $$m = 1.66 \times 10^{-27} \mathrm{~kg}$$ (mass of the proton) and $$v = 3.00 \times 10^7 \mathrm{~m/s}$$.

So,

$$p = (1.66 \times 10^{-27} \mathrm{~kg}) \times (3.00 \times 10^7 \mathrm{~m/s}) = 4.98 \times 10^{-20} \mathrm{~kg \cdot m/s}$$

Next, we find the uncertainty in momentum ($$\Delta p$$) using the given accuracy in measuring velocity:

Given that velocity can be measured to an accuracy of $$\pm 2\%$$:

$$\Delta v = 0.02 \times 3.00 \times 10^7 \mathrm{~m/s} = 6.00 \times 10^5 \mathrm{~m/s}$$

So,

$$\Delta p = m \Delta v = (1.66 \times 10^{-27} \mathrm{~kg}) \times (6.00 \times 10^5 \mathrm{~m/s}) = 9.96 \times 10^{-22} \mathrm{~kg \cdot m/s}$$

We now substitute $$\Delta p$$ into the Heisenberg Uncertainty Principle:

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

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

$$\Delta x \geq \frac{1.05 \times 10^{-34} \mathrm{~Js}}{2 \times 9.96 \times 10^{-22} \mathrm{~kg \cdot m/s}}$$

$$\Delta x \geq \frac{1.05 \times 10^{-34}}{1.992 \times 10^{-21}} \mathrm{~m}$$

$$\Delta x \geq 5.27 \times 10^{-14} \mathrm{~m}$$

This calculation gives the uncertainty in the proton's position as approximately $$5.27 \times 10^{-14} \mathrm{~m}$$, which closely matches option B:

Option B: $$5.29 \times 10^{-14} \mathrm{~m}$$

Thus, the answer is Option B.

Atomic Structure

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