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Step-by-Step Solution
Step 1: State the Heisenberg Uncertainty Principle
According to the Heisenberg uncertainty principle, the product of the uncertainties in position ($\Delta x$) and momentum ($\Delta p$) is constrained by
$ \Delta x \, \Delta p \ge \frac{h}{4\pi} $.
Step 2: Express the Uncertainty in Position
Momentum $p$ can be written as $p = m \, v$. Therefore, the uncertainty in momentum is $\Delta p = m \,\Delta v$. Substituting in the uncertainty principle gives:
$ \Delta x = \frac{h}{4 \pi \, m \,\Delta v} $.
Step 3: Relate Speed to Temperature using Kinetic Theory
For an ideal gas particle of mass $m$ at temperature $T$, a typical (rms) speed is given by:
$ v_{\mathrm{rms}} = \sqrt{\frac{3 k_B T}{m}} $.
Hence, we can take $ \Delta v \approx \sqrt{\frac{3 k_B T}{m}} $.
Step 4: Find the Ratio of Uncertainties for Electron and Proton
Let $m_e$ be the mass of an electron and $m_p$ be the mass of a proton. Then the uncertainties in their positions are:
$ \Delta x_e = \frac{h}{4\pi\,m_e\,\Delta v_e}, \quad \Delta x_p = \frac{h}{4\pi\,m_p\,\Delta v_p} $.
Using $ \Delta v_e \approx \sqrt{\frac{3 k_B T}{m_e}} $ and $ \Delta v_p \approx \sqrt{\frac{3 k_B T}{m_p}} $, we get:
$ \Delta x_e = \frac{h}{4\pi\,m_e\,\sqrt{\frac{3 k_B T}{m_e}}}
\quad\text{and}\quad
\Delta x_p = \frac{h}{4\pi\,m_p\,\sqrt{\frac{3 k_B T}{m_p}}}. $
Taking the ratio:
$ \frac{\Delta x_e}{\Delta x_p}
= \frac{\frac{h}{4\pi \, m_e \,\sqrt{\frac{3 k_B T}{m_e}}}}{\frac{h}{4\pi \, m_p \,\sqrt{\frac{3 k_B T}{m_p}}}}
= \frac{m_p}{m_e} \times \sqrt{\frac{m_e}{m_p}}
= \sqrt{\frac{m_p}{m_e}}. $
Step 5: Conclude the Correct Ratio
Therefore, the ratio of the uncertainty in determining the position of an electron to that of a proton is
$ \sqrt{\frac{m_p}{m_e}} $.
