Answer:
Minimum uncertainty in velocity of a proton,[tex]\Delta v\ge 3.15\times 10^7\ m/s[/tex]
Explanation:
It is given that,
A proton is confined to a space 1 fm wide, [tex]\Delta x=10^{-15}\ m[/tex]
We need to find the minimum uncertainty in its velocity. We know that the Heisenberg Uncertainty principle gives the uncertainty between position and the momentum such that,
[tex]\Delta p.\Delta x\ge \dfrac{h}{4\pi}[/tex]
Since, p = mv
[tex]\Delta (mv).\Delta x\ge \dfrac{h}{4\pi}[/tex]
[tex]m \Delta v.\Delta x\ge \dfrac{h}{4\pi}[/tex]
[tex]\Delta v\ge \dfrac{h}{4\pi m\Delta x}[/tex]
[tex]\Delta v\ge \dfrac{6.63\times 10^{-34}}{4\pi \times 1.67\times 10^{-27}\times 10^{-15}}[/tex]
[tex]\Delta v\ge 3.15\times 10^7\ m/s[/tex]
So, the minimum uncertainty in its velocity is greater than [tex]3.15\times 10^7\ m/s[/tex]. Hence, this is the required solution.
