Question

An electron moves around protons (nucleus) in a circle of radius 'r'. Assuming that the uncertainty of momentum of electron is of the same order as the momentum itself, the momentum of electron will be:

• A. $\frac{h}{4\pi r}$
• B. $\frac{h}{2\pi r}$
• C. $4\pi hr$
• D. $2\pi hr$

Answer 1

The correct option is A $\frac{h}{4\pi r}$

According to Heisenberg's uncertainty principle:

$\Delta x.\Delta p=\frac{h}{4\pi }$

$\Delta x=$ uncertainty in position

$\Delta p=$ uncertainty in momentum

For an orbit of radius $r$, $\Delta x=r$

Therefore, $\Delta p=\frac{h}{4\pi r}=p$ (as order is same)

Option A is correct