Question

An electron is trapped in a 100 μm cavity. What is its minimum velocity?

Answer 1

Answer 2

To find the minimum velocity of the electron trapped in a 100 μm cavity, we can use the concept of uncertainty principle in quantum mechanics.

The uncertainty principle states that there is a fundamental limit to the precision with which certain pairs of physical properties of a particle, such as position (Δx) and momentum (Δp), can be known simultaneously. Mathematically, it is expressed as:

Δx * Δp >= ħ/2

Where: Δx = Uncertainty in position (in meters) Δp = Uncertainty in momentum (in kg m/s) ħ (h-bar) = Reduced Planck's constant ≈ 1.054571 x 10^-34 J s

In our case, the electron is trapped in a 100 μm cavity, which means Δx = 100 μm = 100 x 10^-6 m.

Now, we want to find the minimum velocity of the electron, which corresponds to its minimum momentum (Δp). To find Δp, we rearrange the uncertainty principle equation:

Δp >= ħ / (2 * Δx)

Substitute the values:

Δp >= 1.054571 x 10^-34 J s / (2 * 100 x 10^-6 m)

Calculate Δp:

Δp >= 5.272855 x 10^-34 kg m/s

Now, since momentum (p) is related to velocity (v) by the equation p = m * v, where m is the mass of the electron, we can find the minimum velocity (v_min) by rearranging the equation:

v_min = Δp / m

The mass of the electron (m) is approximately 9.109 x 10^-31 kg.

Calculate v_min:

v_min = 5.272855 x 10^-34 kg m/s / 9.109 x 10^-31 kg

v_min ≈ 0.0057948 m/s

So, the minimum velocity of the electron trapped in a 100 μm cavity is approximately 0.0057948 meters per second.

Answer 3

To find the minimum velocity of an electron trapped in a 100 μm cavity, we can use the Heisenberg Uncertainty Principle, which states that there is a fundamental limit to how precisely we can simultaneously know the position and momentum of a particle.

The Heisenberg Uncertainty Principle is given by:

Δx * Δp ≥ h / (4 * π)

where: Δx = uncertainty in position Δp = uncertainty in momentum h = Planck's constant = 6.626 x 10^-34 J s π = pi (approximately 3.14159)

In this case, the electron is trapped in a cavity with a size of 100 μm. Let's assume that the uncertainty in the position of the electron (Δx) is approximately equal to the size of the cavity.

Δx = 100 μm = 100 x 10^-6 meters

Now, we need to find the minimum uncertainty in momentum (Δp) to calculate the minimum velocity. Since the electron is confined in a small region, it behaves more like a particle with well-defined momentum than a wave.

Δp = h / (4 * π * Δx)

Δp = 6.626 x 10^-34 J s / (4 * π * 100 x 10^-6 meters)

Δp ≈ 5.27 x 10^-26 kg m/s

Now, to find the minimum velocity (v_min) of the electron, we use the equation:

v_min = Δp / m

where m is the mass of the electron, approximately 9.109 x 10^-31 kg.

v_min = (5.27 x 10^-26 kg m/s) / (9.109 x 10^-31 kg)

v_min ≈ 5.79 x 10^4 m/s

So, the minimum velocity of the electron trapped in the 100 μm cavity is approximately 5.79 x 10^4 meters per second.