An electron is trapped in a quantum well that is 48.6 nm wide.Calculate the minimum uncertainty in its velocity. A)7.49 × 10³ m/s B)1.79 × 10⁴m/s C)2.98 × 10⁴ m/s D)3.34 × 10⁴ m/s

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The minimum uncertainty in the electron's velocity is given by the Heisenberg uncertainty principle:$$\Delta x \Delta p \geq \frac{h}{4\pi}$$where $\Delta x$ is the uncertainty in the electron's position, $\Delta p$ is the uncertainty in the electron's momentum, and $h$ is the Planck constant.In this case, the electron is trapped in a quantum well that is 48.6 nm wide, so the uncertainty in its position is $\Delta x = 48.6 ext{ nm} = 48.6 imes 10^{-9} ext{ m}$.The uncertainty in the electron's momentum is related to the uncertainty in its velocity by the equation:$$\Delta p = m \Delta v$$where $m$ is the mass of the electron and $\Delta v$ is the uncertainty in the electron's velocity.Substituting the values of $\Delta x$ and $\Delta p$ into the Heisenberg uncertainty principle, we get:$$48.6 imes 10^{-9} ext{ m} imes m \Delta v \geq \frac{6.626 imes 10^{-34} ext{ J s}}{4\pi}$$Solving for $\Delta v$, we get:$$\Delta v \geq \frac{6.626 imes 10^{-34} ext{ J s}}{4\pi imes 48.6 imes 10^{-9} ext{ m} imes m}$$Substituting the mass of the electron, $m = 9.11 imes 10^{-31} ext{ kg}$, we get:$$\Delta v \geq 7.49 imes 10^3 ext{ m/s}$$Therefore, the minimum uncertainty in the electron's velocity is 7.49 × 10³ m/s.

An Electron Is Trapped in a Quantum Well That Is