If the maximum velocity v of an electron in copper wire is about 2.0x10^8 cm/s, with an uncertainty ≤ 1%, how accurately can the electron's position be determined? In other words, what is the uncertainty in the position of the electron (in meters)?

Question

If the maximum velocity v of an electron in copper wire is about 2.0x10^8 cm/s, with an uncertainty ≤ 1%, how accurately can the electron's position be determined? In other words, what is the uncertainty in the position of the electron (in meters)?

Answer 1

In one second it is 2E8 cm x 0.01 = ? and convert that answer in cm to meters.

Answer 2

To determine the uncertainty in the position of the electron, we can use the Heisenberg's uncertainty principle, which states that the uncertainty in the position and momentum of a particle is related by the equation:

Δ𝑥 ⋅ Δ𝑝 ≥ ℎ/4π

where Δ𝑥 is the uncertainty in position, Δ𝑝 is the uncertainty in momentum, ℎ is Planck's constant (6.626 x 10^(-34) J⋅s), and π is a mathematical constant approximately equal to 3.14159.

In this case, the uncertainty in momentum can be calculated using the velocity and mass of the electron. The mass of an electron is approximately 9.109 x 10^(-31) kg. The momentum of an electron is given by the equation:

𝑝 = 𝑚⋅𝑣

where 𝑝 is the momentum, 𝑚 is the mass, and 𝑣 is the velocity.

First, convert the velocity to meters per second:

𝑣 = 2.0 x 10^8 cm/s = 2.0 x 10^6 m/s

Now, calculate the uncertainty in the momentum:

Δ𝑝 = (1% of 𝑝) = 0.01𝑝 = 0.01⋅𝑚⋅𝑣

Next, calculate the uncertainty in position:

Δ𝑥 ⋅ Δ𝑝 ≥ ℎ/4π

Δ𝑥 ⋅ 0.01⋅𝑚⋅𝑣 ≥ ℎ/4π

Δ𝑥 ≥ ℎ/(4π⋅0.01⋅𝑚⋅𝑣)

Finally, plug in the values and calculate the uncertainty in position:

Δ𝑥 ≥ (6.626 x 10^(-34) J⋅s)/(4π⋅0.01⋅9.109 x 10^(-31) kg⋅2.0 x 10^6 m/s)

Δ𝑥 ≥ 1.445 x 10^(-5) meters

Therefore, the uncertainty in the position of the electron is approximately 1.445 x 10^(-5) meters.