Assume the Heisenberg uncertainty principle can take the form ΔEΔt ≥ ħ. How accurate can the position of an electron be made if its speed is 5 × 10^6 m/s and if the uncertainty in... Assume the Heisenberg uncertainty principle can take the form ΔEΔt ≥ ħ. How accurate can the position of an electron be made if its speed is 5 × 10^6 m/s and if the uncertainty in its energy is 10 eV? Read more

Steps to Solve

1. Identify the uncertainty equation We use the Heisenberg uncertainty principle in the form: $$ \Delta E \Delta t \geq \hbar $$ Where $\Delta E$ is the uncertainty in energy and $\Delta t$ is the uncertainty in time.

2. Convert energy uncertainty to Joules The uncertainty in energy (ΔE) is given as 10 eV. We need to convert this to Joules using the conversion factor: $$ 1 \text{ eV} = 1.6 \times 10^{-19} \text{ J} $$ So, $$ \Delta E = 10 \text{ eV} \times 1.6 \times 10^{-19} \text{ J/eV} = 1.6 \times 10^{-18} \text{ J} $$

3. Calculate uncertainty in time Using the uncertainty in energy, we can find the uncertainty in time (Δt): $$ \Delta t \geq \frac{\hbar}{\Delta E} $$ Where $\hbar$ (the reduced Planck constant) is $1.055 \times 10^{-34} \text{ J s}$. Thus, $$ \Delta t \geq \frac{1.055 \times 10^{-34} \text{ J s}}{1.6 \times 10^{-18} \text{ J}} $$

4. Simplify the calculation Calculating Δt gives: $$ \Delta t \geq 6.59 \times 10^{-17} \text{ s} $$

5. Calculate uncertainty in position (Δx) Using the relationship between speed, uncertainty in position, and uncertainty in time: $$ \Delta x \approx v \Delta t $$ Where $v = 5 \times 10^6 \text{ m/s}$. Therefore, $$ \Delta x \approx 5 \times 10^6 \text{ m/s} \times 6.59 \times 10^{-17} \text{ s} $$

6. Final calculation Calculate Δx: $$ \Delta x \approx 3.295 \times 10^{-10} \text{ m} = 0.3295 \text{ nm} $$