How accurately can the position of a 2.7keV electron be measured?

[ronpaulkid]

Homework Statement

How accurately can the position of a 2.70keV electron be measured assuming its energy is known to 3.00%?

The answer is (deltaX)>=?

Homework Equations

(deltaX)(deltaP)>=h/2pi

(deltaX)(deltaE)>=h/2pi

The Attempt at a Solution

My attempt at this solution was first trying to calculate deltaX using each of the two equations listed above. I was not able to cancel out the units correctly which leads me to believe that the mass of the electron is somehow included (9.11*10^-31 kg).

 

[zhermes]

Your second equation is incorrect; it should be \Delta t \cdot \Delta E \ge \frac{h}{2 \pi}

Using the first equation, you need to know the uncertainty in momentum, but you're given the uncertainty in energy. How can you convert?

 

[cupid.callin]

use the 2.7 keV energy to find the speed of electron

now find \Delta v

you know mass and \Delta v of electron now

just use eqn 1

and the best way to save yourself form unit's trouble is to use SI units ... that way \Delta x will be in metres (m)