The uncertainty in the position of a proton is 250 pm. Estimate the kinetic energy of the proton. A) 1.3 × 10⁻²³ eV B) 8.3 × 10⁻⁵ eV C) 0.013 eV D) 0.24 eV E) 100 MeV

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Quizplus · Accepted Answer

The uncertainty in the position ($\Delta x$) of a proton can be used to estimate the uncertainty in its momentum ($\Delta p$) using the Heisenberg uncertainty principle, $\Delta x \Delta p \geq \frac{\hbar}{2}$, where $\hbar$ is the reduced Planck's constant ($\approx 1.0545718 \times 10^{-34} m^2 kg / s$). Given $\Delta x = 250 pm = 250 \times 10^{-12} m$, we can find $\Delta p$. The kinetic energy ($K$) can then be estimated using $K = \frac{(\Delta p)^2}{2m}$, where $m$ is the mass of the proton ($\approx 1.6726219 \times 10^{-27} kg$). Following through with these calculations leads to an answer in the ballpark of $10^{-24}$ to $10^{-23}$ Joules, which, when converted to electronvolts (1 eV = $1.602 \times 10^{-19}$ Joules), gives a value in the range of $10^{-5}$ eV, making option B the correct choice.

The Uncertainty in the Position of a Proton Is 250