Question

An electron is confined within a region of width 1.0x10^-10 m. (a) Estimate the minimum uncertainty in x - component of electron's momentum (b) if the electron has momentum with magnitude equal to the uncertainty found in part (a) what is its K.E.? Mass of electron = 9.1x10^-31kg.​

Answer 1

Explanation:

part

(a) what is its K.E.?

Mass of electron = 9.1x10^-31kg.

Expert's answer

(a) The Heisenberg's uncertainty principle says

\Delta p\cdot \Delta x\geq \hbarΔp⋅Δx≥ℏ

Hence,

p\sim\Delta p=\frac{\hbar}{\Delta x}\\ =\frac{1.05\times 10^{-34}\:\rm J\cdot s}{1.0\times 10^{-10} \:\rm m}=1.05\times 10^{-24}\:\rm kg\cdot m/sp∼Δp=Δxℏ=1.0×10−10m1.05×10−34J⋅s=1.05×10−24kg⋅m/s

(b)

E_k=\frac{p^2}{2m}=\frac{(1.05\times 10^{-24}\:\rm kg\cdot m/s)^2}{2\times 9.1\times 10^{-31}\:\rm kg}=6.1\times 10^{-19}\:\rm JEk=2mp2=2×9.1×10−31kg(1.05×10−24kg⋅m/s)2=6.1×10−19J