A nonrelativistic proton is confined to a length of 2.0 pm on the x-axis.What is the kinetic energy of the proton if its speed is equal to the minimum uncertainty possible in its speed? (1 eV = 1.60 × 10^-19 J,h = 6.626 × 10^-34J ∙ s,m_proton = 1.67 × 10^-27 kg) A)5.1 eV B)51 eV C)510 eV D)5100 eV E)51,000 eV

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Quizplus · Accepted Answer

The uncertainty principle, $\Delta x \Delta p \geq \frac{h}{4\pi}$, where $\Delta x$ is the uncertainty in position (2.0 pm = 2.0 × 10^-12 m) and $\Delta p$ is the uncertainty in momentum, can be used to find the minimum uncertainty in the proton's momentum. The minimum uncertainty in speed, $\Delta v$, can be related to the minimum uncertainty in momentum by $\Delta p = m_{proton} \Delta v$. Solving for $\Delta v$ gives $\Delta v = \frac{h}{4\pi m_{proton} \Delta x}$. The kinetic energy, $K$, is given by $K = \frac{1}{2} m_{proton} v^2$, but using the minimum uncertainty in speed, we use $\Delta v$ for $v$, so $K = \frac{1}{2} m_{proton} (\Delta v)^2$. Substituting the expression for $\Delta v$ and the given values (with $h = 6.626 \times 10^{-34} \, \text{J} \cdot \text{s}$ and $m_{proton} = 1.67 \times 10^{-27} \, \text{kg}$) yields a kinetic energy in the range of tens of eV, specifically around 51 eV after calculation.

A Nonrelativistic Proton Is Confined to a Length of 2