show that an electron confined to a box of nuclear dimensions ( ) must have minimum energy 20Mev.
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The dimension is 4.9 m
To find:
the dimension of the box
Given:
the minimum energy is 20MeV
Solution:
According to Heisenberg's uncertainty principle, it is impossible to precisely measure or quantify an object's position or momentum. The wave-particle duality of the matter is the basis for this idea.
Using Heisenberg's relation of uncertainty for position and momentum
Δx Δp ≥ℏ/2
Δpc = ℏc/2Δx = 20MeV
where Δx is the width of the region.
ℏc/2Δx = 20MeV
Planck's constant h = 6:6261 * 10^34 J s
speed of light c = 2:9979 * 10^8 m s^-1
hc = 1239.8 eV nm = 1239.8 keV pm = 1239:8 MeV fm
where
eV = 1:6022 * 10^19 J
Also,
ℏc = 197.32 eV nm = 197.32 keV pm = 197.32 MeV fm
where ℏ= h/2π.
ℏc/ 20MeV*2 = Δx
197.38 / 40 = 4.9 m
Δx= 4.9 m
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