# Heisenberg's uncertainty principle question

## Homework Statement

The position and momentum of a 1.00 KeV electron are simultaneously determined. If the position is located to within 0.100 nm, what is the percentage of uncertainty in its momentum?

(Arthur Beiser - Concepts of modern Physics, 3rd Part exercise 33)

## Homework Equations

$$Δx\cdotΔp≥\frac{\hbar}{2}$$

## The Attempt at a Solution

$$Δx\cdotΔp≥\frac{\hbar}{2}\rightarrow \; Δp≥ \frac{\hbar}{2\cdotΔx}$$
Given that Δx=0,100 nm, we can evaluate the uncertainty in its momentum Δp.
The momentum of the particle should be of roughly equal magnitude, and evaluating the fraction $$\frac{Δp}{p}%$$ gives me 98%, where the answer is 3,1 %. What did I do wrong?

MathematicalPhysicist
Gold Member
What did you plug for p?

It should be p=1[Kev] /c where c is the speed of light.

Yep.

$$K=\sqrt{m_{0}^{2}c^{4}+p^{2}c^{2}}≈pc$$
so
$$p=\frac{1,6\cdot10^{-16} J}{3\cdot10^{8} \frac{m}{s}}=0,533 \cdot 10^{-24} kg\cdot\frac{m}{s}$$
and because $$Δp=\frac{1,05\cdot10^{-34}J\cdot s}{2\cdot 10^{-10}m}=0,525\cdot 10^{-24} kg\cdot \frac{m}{s}$$
we get
$$\frac{Δp}{p}≈0,98$$

:(

Could I get a bumpity-bump 'cause these exercises are crucial for the finals?