this is a question i answered about basic knowledge of Heisenberg's principle. but i also tried to give the asker a good background on the subject and even a nice illustration of the meaning of the results at the end.
Q:
At a baseball game, a radar gun measures the speed of a 140 g baseball to be 137.32 ± 0.05 km/h.
(a) What is the minimum uncertainty of the position of the baseball?
answer in m
(b) If the speed of a proton is measured to the same precision, what is the minimum uncertainty in its position?
answer in m
please help me out i'm stuck it would be greatly apprieciated if you help me out..thanks
A:
a ) here you need the relationship between uncertainty in position and uncertainty in speed. unfortunately, no such one exists, but there's an equally useful one which relates uncertainty in position to uncertainty in momentum. it is one of Heisenberg's equations:
∆x * ∆p ≥ ħ/2
(where ħ is "h-bar" or h/2π)
(and ∆ means "uncertainty in")
it is good then that you are given the mass of the ball, so that p can be written as mv, with those 2 values given in the problem.
now, all that is left to do is solve for ∆x (or the uncertainty in position):
∆x ≥ ħ / (2*∆mv)
note that since there is no uncertainty in the mass, we can take it out of the ∆ operator - not a necessary step, but still worth noting.
∆x ≥ ħ / (2m*∆v)
so now all we have to do is plug in our known values for m, ∆v, and ħ (in the correct mks units, of course)
m = 140 g = 0.140 kg
∆v = 0.05 km/hr = 50 m/hr = 0.0139 m/s
ħ = 1.0546 * 10^-34 J*s = 1.0546 * 10^-34 kg*m²/s
thus we have:
∆x ≥ (1.0546 * 10^-34 kg*m²/s) / (2 * 0.140 kg *0.0139 m/s)
cancelling units and evaluating, we get:
∆x ≥ 2.71 * 10^-32 meters <<<<<<<<<<<<<<< (answer part a)
you'll notice that this is so small, it is not even perceptible - or anywhere near that! that is why you don't notice this phenomenon while watching a game of baseball.
b) for this problem, we need the same uncertainty relationship. ∆v is again 0.05 km/hr (or 0.0139 m/s). and the mass of the proton can be looked up and is approximately 1.6726 * 10^-27 kg (much much smaller than the baseball)
starting again with the basic relationship
∆x * ∆p ≥ ħ/2, and rearranging:
∆x ≥ ħ / (2m*∆v)
plugging in values exactly the same as last time:
∆x ≥ (1.0546 * 10^-34 kg*m²/s) / (2 * 1.6726 * 10^-27 kg * 0.0139 m/s)
again, all we are left with is meters, and the value of the expression is
∆x ≥ 2.269 * 10^-6 meters <<<<<<<<<<< (answer part b)
although this is still relatively small, it is many many many orders of magnitudes larger than the ∆x of the baseball.
consider also that it is roughly 2.27 * 10^-6 meters (2 micrometers). it is interesting to see this value compared to the approximate size of of an atom, which has a diameter of roughly 100 PICOmeters (a picometer is 10^-12 meters. or 100 of them is 10^-10 meters).
so the uncertainty in the position of this moving proton is almost 4000 times larger than the diameter of an atom!! so forget about it if you wanted to pinpoint the location of this particular proton among other atoms.
i hope this was helpful! and i'm always happy to answer more questions :]
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