Using Werner Hesienbergs uncertainty principle
delta x >h/ (4pie*m*delta v)
when mass of electron is 9.11x10^-31kg
what is the uncertainty in the position of an electron moving at 9x10^6 m/s with an uncertainty of delta v= 0.01x10^6m/s
To find the uncertainty in the position of an electron using Heisenberg's uncertainty principle, we can use the formula:
Δx > h / (4πmΔv),
where Δx represents the uncertainty in position, h is Planck's constant (6.626 x 10^-34 J·s), m is the mass of the electron (9.11 x 10^-31 kg), and Δv is the uncertainty in velocity.
Given that the velocity of the electron is 9 x 10^6 m/s and the uncertainty in velocity is 0.01 x 10^6 m/s, we can substitute these values into the formula to calculate the uncertainty in position:
Δx > (6.626 x 10^-34 J·s) / (4π * 9.11 x 10^-31 kg * 0.01 x 10^6 m/s).
First, calculate the denominator:
(4π * 9.11 x 10^-31 kg * 0.01 x 10^6 m/s) = 4π * (9.11 x 0.01) x (10^-31 kg x 10^6 m/s)
= 36.44π x 10^-25 kg·m/s.
Now, substitute the values into the formula:
Δx > (6.626 x 10^-34 J·s) / (36.44π x 10^-25 kg·m/s).
Calculate the numerator:
(6.626 x 10^-34 J·s) = 0.6626 x 10^-33 J·s.
Finally, divide the numerator by the denominator:
Δx > (0.6626 x 10^-33 J·s) / (36.44π x 10^-25 kg·m/s)
≈ 2.28 x 10^-10 kg·m/s.
Therefore, the uncertainty in the position of the electron is approximately 2.28 x 10^-10 m.