How accurately can the position of an electron be determined if its speed is known and its energy is 10 eV, according to the Heisenberg uncertainty principle?

Steps to Solve

1. Recall the Heisenberg Uncertainty Principle

The Heisenberg uncertainty principle states that the uncertainty in position $\Delta x$ and the uncertainty in momentum $\Delta p$ of a particle are related by the equation:

$$ \Delta x \Delta p \geq \frac{\hbar}{2} $$

where $\hbar$ (reduced Planck's constant) is approximately $1.055 \times 10^{-34} \text{ Js}$.

2. Calculate the momentum uncertainty

The momentum $p$ is given by $p = mv$, where $m$ is the mass of the electron ($m \approx 9.11 \times 10^{-31} \text{ kg}$) and $v$ is its speed. The kinetic energy $K.E.$ can also be expressed as:

$$ K.E. = \frac{1}{2}mv^2 $$

Given the energy $K.E. = 10 \text{ eV}$, we convert it to Joules:

$$ 10 \text{ eV} = 10 \times 1.6 \times 10^{-19} \text{ J} = 1.6 \times 10^{-18} \text{ J} $$

Rearranging to find $v$:

$$ v = \sqrt{\frac{2 K.E.}{m}} = \sqrt{\frac{2 \times 1.6 \times 10^{-18}}{9.11 \times 10^{-31}}} $$

Calculating $v$ gives the speed of the electron.

3. Determine the uncertainty in momentum

Using the speed obtained:

$$ p = mv $$

The uncertainty in momentum $\Delta p$ is approximated as:

$$ \Delta p \approx m \Delta v $$

Since we assume small velocities for the uncertainty, we will consider $\Delta v$ as a small fraction of $v$.

4. Find the uncertainty in position

Now, insert $\Delta p$ back into the Heisenberg Uncertainty Principle equation:

$$ \Delta x \geq \frac{\hbar}{2 \Delta p} $$

Given that $\Delta p = m \Delta v$, we can express the position uncertainty as:

$$ \Delta x \geq \frac{\hbar}{2 m \Delta v} $$

Substituting known values will give us the uncertainty in position.