An electron is confined to a region of the size of an atom (0.1 nm). What is the minimum uncertainty of the momentum of the electron? What is the kinetic energy of an electron with a momentum equal to this uncertainty?

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Accepted Answer

Using the Uncertainty Principle, the minimum uncertainty of the momentum of an electron in the atom can be written
[latex]\Delta p=\frac{\hbar }{2\Delta x}=\frac{1.054573 × 10^{−34} \ J ·s}{2.0 × 10^{−10} \ m}= 5.27 × 10^{-25} \ kg m/s .[/latex]
In order to calculate the kinetic energy of an electron with this momentum, we write the kinetic energy in terms of the momentum as before to obtain
[latex]KE =\frac{1}{2m}p^{2}.[/latex]
Substituting the value of the uncertainty of the momentum in place of the momentum in the above equation gives
[latex]K E=\frac{(5.27 × 10^{−25} \ kg m/s)^{2}}{2 · 9.11 × 10^{−31} \ kg} imes \frac{1 \ eV}{1.602 × 10^{−19} \ J} = 0.95 \ eV . [/latex]
This is a reasonable estimate of the energy of the outer electrons in an atom. The inner electrons are more tightly bound having smaller uncertainties in their position and corresponding greater uncertainties in their momentum.

Question 3.3: An electron is confined to a region of the size of an atom (...

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