An electron is confined inside a cubical box 1.0×10^-10 ms on a side.
1. Estimate the minimum uncertainty in the electrons momentum .
2. If the electron has momentum with magnitude equals to the uncertainty found in 1, given that the mass of an electron is 9.11×10^-31kg.what is its kinetic energy? Express the energy in joules.
1. The minimum uncertainty in the momentum of the electron can be estimated using the Heisenberg uncertainty principle, which states that the product of the uncertainty in position and momentum is greater than or equal to h/4π, where h is the reduced Planck's constant (h/2π).
The uncertainty in position can be approximated as half the length of the side of the box, so Δx = 0.5 × 1.0×10^-10 m = 5.0×10^-11 m.
Using the Heisenberg uncertainty principle, Δp * Δx >= h/4π, we can solve for the minimum uncertainty in momentum, Δp:
Δp * 5.0×10^-11 m >= (h/2π)
Δp >= h/(2π * 5.0×10^-11 m)
Δp >= (6.63×10^-34 J s) / (2π * 5.0×10^-11 m)
Δp >= 2.11×10^-21 kg m/s
Therefore, the minimum uncertainty in the electron's momentum is approximately 2.11×10^-21 kg m/s.
2. Since the electron's momentum with magnitude equals to the uncertainty found in 1, we can consider the momentum to be Δp = 2.11×10^-21 kg m/s.
The kinetic energy of the electron can be calculated using the formula: KE = p^2 / (2m), where p is momentum and m is the mass of the electron.
Substitute the values:
KE = (2.11×10^-21 kg m/s)^2 / (2 * 9.11×10^-31 kg)
KE = (4.4521×10^-42) / (1.822×10^-30)
KE = 2.443×10^-12 J
Therefore, the kinetic energy of the electron is approximately 2.443×10^-12 Joules.