Question

An electron moving in the positive \(x\) -direction passes through a slit of width \(\Delta y=85 \mathrm{nm} .\) What is the minimum uncertainty in the electron's velocity in the \(y\) -direction?

Short answer

Answer: The minimum uncertainty in the electron's velocity in the y-direction is approximately \(6.75 \times 10^5\text{ m/s}\).

Step-by-step solution

Understanding Heisenberg's Uncertainty Principle

Heisenberg's Uncertainty Principle states that there is a fundamental limit to the precision with which certain pairs of physical properties of a particle can be known simultaneously. In this case, we will use the principle for position and momentum. The formula for Heisenberg's Uncertainty Principle is: \(h \leq 4\pi\Delta y\Delta p_y\) where \(h\) is Planck's constant, \(\Delta y\) is the uncertainty in position (width of the slit), and \(\Delta p_y\) is the uncertainty in momentum in the y-direction. Since the electron's motion is being confined by the slit in the y-direction, we are given \(\Delta y\). Our goal is to find the minimum uncertainty in the velocity of the electron in the y-direction. To do this, we first need to find \(\Delta p_y\).

Finding the uncertainty in momentum, \(\Delta p_y\)

We will solve for \(\Delta p_y\) using the Heisenberg's Uncertainty Principle formula, given by: \(\Delta p_y \geq \frac{h}{4\pi\Delta y}\) Plugging in the given values, we get: \(\Delta p_y \geq \frac{6.626 \times 10^{-34}\text{ Js}}{4\pi (85 \times 10^{-9}\text{m})}\) Calculating this gives: \(\Delta p_y \geq 6.15 \times 10^{-25} \text{ kg m/s}\)

Finding the minimum uncertainty in the electron's velocity in the y-direction, \(\Delta v_y\)

Recall that momentum (p) is equal to mass (m) multiplied by velocity (v). Using this relationship, we can find the minimum uncertainty in the electron's velocity in the y-direction: \(\Delta v_y = \frac{\Delta p_y}{m}\) where m is the mass of the electron, which is \(9.11 \times 10^{-31}\text{ kg}\). Substitute the values for \(\Delta p_y\) and m into the equation: \(\Delta v_y = \frac{6.15 \times 10^{-25}\text{ kg m/s}}{9.11 \times 10^{-31}\text{ kg}}\) Calculating this gives: \(\Delta v_y \approx 6.75 \times 10^5\text{ m/s}\) So, the minimum uncertainty in the electron's velocity in the y-direction is approximately \(6.75 \times 10^5\text{ m/s}\).