Question

A proton is confined to a space 1 fm wide (about the size of an atomic nucleus). What's the minimum uncertainty in its velocity?

Answer 1

Answer:

Minimum uncertainty in velocity of a proton,$\Delta v\ge 3.15\times 10^7\ m/s$      

Explanation:

It is given that,

A proton is confined to a space 1 fm wide, $\Delta x=10^{-15}\ m$

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,

$\Delta p.\Delta x\ge \dfrac{h}{4\pi}$

Since, p = mv

$\Delta (mv).\Delta x\ge \dfrac{h}{4\pi}$

$m \Delta v.\Delta x\ge \dfrac{h}{4\pi}$

$\Delta v\ge \dfrac{h}{4\pi m\Delta x}$

$\Delta v\ge \dfrac{6.63\times 10^{-34}}{4\pi \times 1.67\times 10^{-27}\times 10^{-15}}$

$\Delta v\ge 3.15\times 10^7\ m/s$

So, the minimum uncertainty in its velocity is greater than $3.15\times 10^7\ m/s$. Hence, this is the required solution.