Answer:
1.98 × 10⁻³³m
Explanation:
It is given that,
Mass of the bullet, m = 27 g = 0.027 kg
Velocity of bullet, v = 800 m/s
The uncertainty in momentum is 0.20%. The momentum of the bullet is given by :
[tex]p=mvp=0.027\times 800 = 21.6\ kg-m/s[/tex]
Uncertainty in momentum is,
[tex]\Delta p=0.2\%\ of\ 21.6\\\Delta p=0.0432[/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}\Delta x=\dfrac{h}{4\pi \Delta p}\Delta x=\dfrac{6.62\times 10^{-34}}{4\pi \times 0.0432}\Delta x=1.98\times 10^{-33}\ m[/tex]
