Heisenberg's uncertainty principle question

[karkas]

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]

What did you plug for p?

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

 

[karkas]

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
$$\frac{Δp}{p}≈0,98$$

:(

 

[karkas]

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

 

[asteriatic]

You did not do anything wrong. The answer of 3.1% is correct. The 98% value you obtained is the percentage of uncertainty in the momentum, not the percentage of the momentum itself. The uncertainty principle states that the product of the uncertainties in position and momentum must be greater than or equal to a certain value, in this case, ħ/2. Therefore, the uncertainty in momentum cannot be equal to 98% of the actual momentum, as that would violate the principle. Instead, the uncertainty in momentum must be at least 3.1% of the actual momentum.