Question

What is the minimum kinetic energy of an electron confined to a region the size of an atomic nucleus \((1.0 \mathrm{fm}) ?\)

Short answer

Answer: The minimum kinetic energy of an electron confined to a region the size of an atomic nucleus (1.0 fm) is approximately 1.763 × 10^(-14) J or 17.63 MeV.

Step-by-step solution

Calculate uncertainty in momentum (∆p)

Using Heisenberg Uncertainty Principle, ∆x * ∆p ≥ h/(4π) Given: ∆x = 1.0 fm = 1.0 × 10^(-15) m We want to find the minimum possible uncertainty in momentum (∆p), so we must find the equality. ∆p = (h/(4π))/ ∆x Now, substitute the known values and the given ∆x value: ∆p = ((6.626 × 10^(-34) Js)/(4π)) / (1.0 × 10^(-15) m)

Calculate the minimum kinetic energy (K.E.)

The formula for kinetic energy is: K.E. = p^2/(2m) The minimum kinetic energy will occur when the momentum (p) equals the minimum possible uncertainty in momentum (∆p). So, we can rewrite the formula as: K.E_min = (∆p)^2 / (2m) Substitute the value of ∆p from Step 1 and the mass of the electron (m = 9.11 × 10^(-31) kg): K.E_min = ((6.626 × 10^(-34) Js / (4π))/(1.0 × 10^(-15) m))^2 / (2 * (9.11 × 10^(-31) kg))

Calculate the numerical value of the minimum kinetic energy

Now, we can plug in the numbers and calculate the minimum kinetic energy. K.E_min = ((6.626 × 10^(-34) Js / (4π))/(1.0 × 10^(-15) m))^2 / (2 * (9.11 × 10^(-31) kg)) K.E_min ≈ 1.763 × 10^(-14) J Thus, the minimum kinetic energy of an electron confined to a region the size of an atomic nucleus (1.0 fm) is approximately 1.763 × 10^(-14) J or 17.63 MeV.