A nonrelativistic electron is confined to a length of 500 pm on the x-axis.What is the kinetic energy of the electron if its speed is equal to the minimum uncertainty possible in its speed? (h = 6.626 × 10^-34 J ∙ s,m_el = 9.11 × 10^-31 kg,1 eV = 1.60 × 10^-19 J) A)0.015 eV B)0.15 eV C)1.5 eV D)15 eV E)150 eV

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The minimum uncertainty in speed, $\Delta v$, can be found using Heisenberg's uncertainty principle, which states $\Delta x \Delta p \geq \frac{h}{4\pi}$, where $\Delta x$ is the uncertainty in position (500 pm = $500 \times 10^{-12}$ m), $h$ is Planck's constant ($6.626 \times 10^{-34}$ J·s), and $\Delta p$ is the uncertainty in momentum. Since $\Delta p = m \Delta v$, we can solve for $\Delta v$.$$\Delta x m \Delta v \geq \frac{h}{4\pi}$$Solving for $\Delta v$ gives:$$\Delta v \geq \frac{h}{4\pi m \Delta x}$$Substituting the given values:$$\Delta v \geq \frac{6.626 \times 10^{-34}}{4\pi (9.11 \times 10^{-31})(500 \times 10^{-12})}$$Calculating $\Delta v$ gives a minimum speed due to the uncertainty. The kinetic energy ($K$) can then be found using $K = \frac{1}{2}mv^2$, where $v = \Delta v$ for minimum uncertainty. Substituting the value of $\Delta v$ into the kinetic energy formula and converting the result to electron volts (eV) using $1 eV = 1.60 \times 10^{-19} J$ gives a value close to 1.5 eV.

A Nonrelativistic Electron Is Confined to a Length of 500