Question

A 50.0 -kg particle has a de Broglie wavelength of \(20.0 \mathrm{~cm} .\). a) How fast is the particle moving? b) What is the smallest uncertainty in the particle's speed if its position uncertainty is \(20.0 \mathrm{~cm} ?\)

Short answer

Question: Calculate the speed and the smallest uncertainty in the speed of a particle with a mass of 50.0 kg and a de Broglie wavelength of 20.0 cm when the position uncertainty is 20.0 cm. Answer: The particle has a speed of \(6.626 \times 10^{-36} \mathrm{m/s}\) and the smallest uncertainty in its speed is \(1.325 \times 10^{-36} \mathrm{~m/s}\).

Step-by-step solution

Recall de Broglie wavelength formula

The de Broglie wavelength formula is given by: \[ \lambda = \frac{h}{p} \] where \(\lambda\) is the wavelength, \(h\) is the Planck's constant, and \(p\) is the momentum.

Rearrange the formula to solve for the momentum

Rearrange the formula to find the momentum, which is: \[ p = \frac{h}{\lambda} \]

Calculate momentum

Now plug in the given values into the formula: \[ p = \frac{6.626 \times 10^{-34} \mathrm{~Js}}{20.0 \times 10^{-2} \mathrm{~m}} \] \[ p = 3.313 \times 10^{-34} \mathrm{~kg~m/s} \]

Use the momentum to find the speed

The formula for momentum is: \[ p = mv \] Rearrange to solve for the speed, \(v\): \[ v = \frac{p}{m} \] Now plug in the values for momentum and mass: \[ v = \frac{3.313 \times 10^{-34} \mathrm{~kg~m/s}}{50.0 \mathrm{~kg}} \] \[ v = 6.626 \times 10^{-36} \mathrm{~m/s} \] So the particle is moving at a speed of \( 6.626 \times 10^{-36} \mathrm{~m/s} \). b) Smallest Uncertainty in the Particle's Speed:

Recall Heisenberg uncertainty principle formula

The Heisenberg uncertainty principle formula is given by: \[ \Delta x \Delta p \geq \frac{h}{4\pi} \] We have the position uncertainty, \(\Delta x\), and we need to find the smallest speed uncertainty, \(\Delta v\), using this formula.

Use momentum formula and uncertainty principle

Since \(p = mv\), we have \(\Delta p = m\Delta v\). Substitute this into the Heisenberg uncertainty principle formula: \[ \Delta x m\Delta v \geq \frac{h}{4\pi} \]

Rearrange the formula to solve for speed uncertainty

Rearrange the formula to solve for the smallest speed uncertainty, \(\Delta v\): \[ \Delta v \geq \frac{h}{4\pi m\Delta x} \]

Calculate speed uncertainty

Now plug in the given values into the formula: \[ \Delta v \geq \frac{6.626 \times 10^{-34} \mathrm{~Js}}{4\pi \times 50.0 \mathrm{~kg} \times 20.0 \times 10^{-2} \mathrm{~m}} \] \[ \Delta v \geq 1.325 \times 10^{-36} \mathrm{~m/s} \] So the smallest uncertainty in the particle's speed is \(1.325 \times 10^{-36} \mathrm{~m/s}\).