Question

The uncertainty in position of a proton confined to the nucleus of an atom is roughly the diameter of the nucleus. If this diameter is 7.8x10^-15 m, what is the uncertainty in the proton's momentum? Express your answer using two significant figures.

Answer 1

Answer: The uncertainty in proton's momentum is $1.3\times 10^{-20}kg.m/s$

Explanation:

The equation representing Heisenberg's uncertainty principle follows:

$\Delta x.\Delta p\geq \frac{h}{2\pi}$

where,

$\Delta x$ = uncertainty in position = d = $7.8\times 10^{-15}m$

$\Delta p$ = uncertainty in momentum = ?

h = Planck's constant = $6.627\times 10^{-34}kgm^2/s^2$

Putting values in above equation, we get:

$\Delta p=\frac{6.627\times 10^{-34}kgm^2/s^2}{2\times 3.14\times 7.8\times 10^{-15}m}\\\\\Delta p=1.3\times 10^{-20}kg.m/s$

Hence, the uncertainty in proton's momentum is $1.3\times 10^{-20}kg.m/s$