Minimum energy of electron bound in nucleus

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SUMMARY

The discussion centers on estimating the minimum energy of an electron confined within a nucleus of radius 4 femtometres. Utilizing the Heisenberg uncertainty principle, represented by the equation ΔxΔp_x=h/4π, the average momentum is derived. The kinetic energy of the electron is calculated using the equation E=pc, where the rest energy is negligible compared to kinetic energy. Clarification is sought regarding the use of Δp_x/2 in the calculations, which is explained as a method to represent momentum uncertainty effectively.

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Homework Statement



An electron is confined to a nucleus of radius 4 femtometres. Estimate its minimum energy.

Homework Equations



ΔxΔp_x=h/4\pi
E^2=p^2c^2 + m^2c^4

As the electron's rest energy will be much less than it's kinetic energy,

E=pc

The Attempt at a Solution



So I understand that I'm supposed to use the uncertainty principle to find the average momentum, and then plug it into the energy equation. What I don't understand is that the solutions use Δpx/2 instead of just Δpx. Why is this?
 
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I think that Δpx/2 in both directions gives a difference of Δpx, which is "somehow" equivalent to a momentum uncertainty.
Using Δpx is probably fine, too.
 

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