1) If you confine an electron to a box and know that the uncertainty in the elec
ID: 1411131 • Letter: 1
Question
1) If you confine an electron to a box and know that the uncertainty in the electron's speed is 65 m/s, what is the smallest length that the box could have? (melectron = 9.11 × 10-31 kg, h = 6.626 × 10-34 J s)
2) The radius of a typical nucleus is about 5 × 10-15 m. Assuming this to be the uncertainty in the position of a proton in the nucleus, estimate the uncertainty in the proton's energy, in MeV. (mproton = 1.67 × 10-27 kg, h = 6.626 × 10-34 J s, 1 eV = 1.6 × 10-19 J)
Please show all work. Thank you
Explanation / Answer
1)
?x ?p ? h/4?
?p = m?V
?x = h / 4?*m?V
?x = (6.626 x 10^-34 J-s) / (4 x 3.14 x 9.1 x 10^-31 kg x 65 m/s)
?x = 8.92 x 10^-7 m
2)
Since the proton is confined to the nucleus, ?x = 5 x 10^-15 m, and from the uncertainty principle,
?p = h/4pi*?x
?x = (6.626 x 10^-34 Js) / 4pi(5 x 10^-15m) = 1.055 x10^-20 kg m/s.
The minimum average momentum of the proton is roughly equal to the uncertainty in the momentum, therefore, the minimum kinetic energy is:
KE = p^2/2m
KE = (?p)^2 / 2m
KE = (1.055 x 10^-20 kg m/s)^2 / 2(1.67 x 10-27 kg)
KE = 0.333 x 10^-13 J
KE = 0.208 ×10^6 eV
KE = 0.208 MeV.
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