Answer:
Uncertainty in position of the bullet is [tex]\Delta x=1.07\times 10^{-33}\ m[/tex]
Explanation:
It is given that,
Mass of the bullet, m = 35 g = 0.035 kg
Velocity of bullet, v = 709 m/s
The uncertainty in momentum is 0.20%. The momentum of the bullet is given by :
[tex]p=mv[/tex]
[tex]p=0.035\times 709=24.81\ kg-m/s[/tex]
Uncertainty in momentum is,
[tex]\Delta p=0.2\%\ of\ 24.81[/tex]
[tex]\Delta p=0.049[/tex]
We need to find the uncertainty in position. It can be calculated using Heisenberg uncertainty principal as :
[tex]\Delta p.\Delta x\geq \dfrac{h}{4\pi}[/tex]
[tex]\Delta x=\dfrac{h}{4\pi \Delta p}[/tex]
[tex]\Delta x=\dfrac{6.62\times 10^{-34}}{4\pi \times 0.049}[/tex]
[tex]\Delta x=1.07\times 10^{-33}\ m[/tex]
Hence, this is the required solution.