A) 3.15 mm
B) 3.15 cm
C) 3.15 m
D) 3.15 km
Correct Answer: D
Solution :
| Let \[\Delta {{x}_{0}}=\]uncertainly in position at time \[t=0\,s,\]so uncertainly in momentum is \[\Delta p\ge \frac{h}{4\pi \Delta {{x}_{0}}}\] |
| As \[v<<c,\]momentum uncertainly is \[\Delta p=\Delta (mv)={{m}_{0}}\Delta v;\] |
| Also; the uncertainly in proton's velocity is \[\Delta v=\frac{\Delta p}{{{m}_{0}}}=\frac{h}{4\pi {{m}_{0}}\Delta {{x}_{0}}}\] |
| So, the distance x that proton covers in time t cannot be known more accurately than |
| \[\Delta x=t\Delta v\ge \frac{ht}{4\pi {{m}_{0}}\Delta {{x}_{0}}}\] |
| At \[t=1s,\]value of \[\Delta x\]is |
| \[\Delta x\ge \frac{1.054\times {{10}^{-\,34}}\times 1}{2(1.672\times {{10}^{-\,27}})\times (1\times {{10}^{-\,11}})}\]\[\ge 3.15\times {{10}^{3}}m\] |
| or \[\Delta x\ge 3.15\,km\] |
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