what is the minimum uncertainty in the position of an electron moving at a speed of 4 x 10^6 m/s ± 1 %?

I keep getting 1.44 x10^-11 m, but the right answer is 1x10^-9 m.

never mind! figured it out. forgot to multiply the velocity to 1%

To calculate the minimum uncertainty in position, we can use Heisenberg's uncertainty principle, which states that the uncertainty in the position of a particle multiplied by the uncertainty in its momentum is greater than or equal to a constant value, h/4π.

The uncertainty in momentum can be calculated using the equation: Δp = mΔv, where Δp is the uncertainty in momentum, m is the mass of the electron, and Δv is the uncertainty in the velocity of the electron.

Given:
Speed of the electron (v) = 4 x 10^6 m/s
Uncertainty in the speed (Δv) = 1%

To calculate the uncertainty in velocity, we need to find 1% of the speed and convert it into m/s:
Δv = (1/100) * 4 x 10^6 m/s = 4 x 10^4 m/s

Now, we can calculate the uncertainty in momentum:
Δp = mΔv

The mass of an electron (m) is approximately 9.11 x 10^-31 kg.
Δp = (9.11 x 10^-31 kg) * (4 x 10^4 m/s) = 3.644 x 10^-26 kg·m/s

To find the minimum uncertainty in position (Δx), we rearrange the uncertainty principle:
Δx = h / (4πΔp)

The value of h is Planck's constant, which is approximately 6.626 x 10^-34 J·s.

Δx = (6.626 x 10^-34 J·s) / (4π * 3.644 x 10^-26 kg·m/s)
Δx ≈ 5.69 x 10^-9 m or 5.69 nm

Therefore, the correct minimum uncertainty in the position of the electron is approximately 5.69 nm, not 1.44 x 10^-11 m as you initially calculated.