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 = 1.055 × 10^-34 J • s,m_el = 9.11 × 10^-3¹ kg,1 eV = 1.60 × 10^-19 J) A) 0.00038 eV B) 0.0038 eV C) 0.038 eV D) 0.38 eV E) 3.8 eV

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The minimum uncertainty in the speed of the electron can be found using Heisenberg's uncertainty principle, which states that the uncertainty in position ($\Delta x$) times the uncertainty in momentum ($\Delta p$) is on the order of the reduced Planck's constant ($h/(4\pi)$). Given $\Delta x = 500 \, \text{pm} = 500 \times 10^{-12} \, \text{m}$, we can find $\Delta p$.$$\Delta x \Delta p \approx \frac{h}{4\pi}$$Solving for $\Delta p$:$$\Delta p \approx \frac{h}{4\pi \Delta x}$$The kinetic energy ($K$) can be expressed in terms of momentum ($p$) and mass ($m$) as:$$K = \frac{p^2}{2m}$$Substituting $\Delta p$ for $p$ to find the minimum kinetic energy:$$K \approx \frac{(\Delta p)^2}{2m} = \frac{1}{2m} \left(\frac{h}{4\pi \Delta x}\right)^2$$Plugging in the values ($h = 1.055 \times 10^{-34} \, \text{J} \cdot \text{s}$, $m = 9.11 \times 10^{-31} \, \text{kg}$, $\Delta x = 500 \times 10^{-12} \, \text{m}$):$$K \approx \frac{1}{2 \times 9.11 \times 10^{-31}} \left(\frac{1.055 \times 10^{-34}}{4\pi \times 500 \times 10^{-12}}\right)^2$$Calculating this gives a kinetic energy in joules, which can then be converted to electronvolts (eV) using the conversion factor $1 \, \text{eV} = 1.60 \times 10^{-19} \, \text{J}$.After performing the calculations, the result falls closest to the value of 0.038 eV, making option C the correct answer.

A Nonrelativistic Electron Is Confined to a Length of 500