Answer:[tex]\Delta P=2.1\times 10^{-20} kg.m/s[/tex]
Explanation:
The uncertainty principle is given by
[tex]\Delta x\cdot \Delta P\geq \frac{h}{2\pi }[/tex]
The uncertainty in particle position is equal to the diameter of the nucleus
i.e.
[tex]\Delta x=5\times 10^{-15}m[/tex]
Therefore Uncertainty in particle momentum is
[tex]\Delta x\cdot \Delta P\geq \frac{h}{2\pi }[/tex]
[tex]\Delta P\geq \frac{h}{2\pi \times \Delta x}[/tex]
[tex]\Delta P=2.1\times 10^{-20} kg.m/s[/tex]