An electron has an uncertainty in its position of 507pm. What is the uncertainty in its velocity?
Your answer is going to be Δv = 1.14×105 m/s
To determine the uncertainty in the velocity of an electron, we need to use Heisenberg's uncertainty principle, which states that the product of the uncertainties in position (Δx) and momentum (Δp) of a particle cannot be smaller than Planck's constant (h) divided by 4π.
Δx * Δp ≥ h/4π
In this case, we have the uncertainty in position (Δx) of the electron as 507 pm. To convert this value to meters, we need to divide it by 10^12, as there are 10^12 picometers in a meter.
Δx = 507 pm = 507 × (10^-12) m = 5.07 × (10^-10) m
Now, we can rearrange the uncertainty principle equation to solve for the uncertainty in momentum:
Δp ≥ (h/4π) / Δx
Plugging in the known values:
Δp ≥ (6.626 × 10^-34 J·s / (4π)) / (5.07 × 10^-10 m)
Simplifying the equation:
Δp ≥ (1.05 × 10^-34 J·s) / (5.07 × 10^-10 m)
Next, we can use the definition of momentum (p = mv) to relate momentum to velocity. Rearranging the equation:
Δv = Δp / m
Where Δv is the uncertainty in velocity and m is the mass of the electron.
The mass of an electron is approximately 9.10938356 × 10^-31 kilograms.
Substituting the known values:
Δv = (1.05 × 10^-34 J·s) / (9.10938356 × 10^-31 kg)
Simplifying the equation:
Δv = 1.15 × 10^4 m/s
Therefore, the uncertainty in the velocity of the electron is approximately 1.15 × 10^4 m/s.