Take the frame of reference at a fixed helium nucleus
m1−helium atom
m2−electron
v1=0
EK=2m2v22
m2v2=m1u1+m2u2
m1u1=m2v2−m2u2
max(u1)=m12v2m2
ΔEKm1=2m1u12
max(ΔEKm1)=m12m22v22
maxpercentage=EKmax(ΔEKm1)∗100=m14m2∗100
The helium nucleus - is made up of two protons and two neutrons.
The mass of a proton is 1836 heavier than the mass of an electron
maxpercentage==4∗1836∗m24m2∗100=0.054%
maxpercentage<0.1%
Proven
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