Calculating Electron Energy with Uncertainty Principle: Homework Example

[v_pino]

Homework Statement

Using the uncertainty principle, estimate the minimum energy in electron volts of an electron confined to a spherical region of radius 0.1nm.

Homework Equations

delta-x * delta-p = h-bar / lamda

delta-y * delta-p = h-bar / lamda

delta-z * delta-p = h-bar / lamda

Energy = 0.5* p^2 / m

p = momentum
m = mass of electron

The Attempt at a Solution

By letting the uncertainty in position (delta-x, delta-y, delta-z) equal to r, I got 2.86eV :

Total energy = 3/8 * (h-bar^2 / r^2 * m)

But the answer should be ~ 10ev.

 

[npupp]

[STRIKE]If the electron is confined in a spherical region, the uncertainty in X is given by \Delta X = X_{2} - X_{1}

So taking the centre of the sphere as 0, you will have 0.1nm and -0.1nm as X_{2} and X_{1} respectively, \DeltaX will then be 0.2nm

Hope this helps ;)[/STRIKE]

Mulling it over a coffee. Brb.

 

[betel]

This will make matters worse by a factor of 4. Also on the right hand side of the uncertainty inequality there should be a 1/2, which would give an additional factor of 4 in the wrong direction. Are you sure the given answer is correct.

 

[npupp]

Sure, no. I'll consider over a coffee.

 

[Deric Boyle]

One can derive the general uncertainity principle by using the standard deviation method of statistics.
Here delta(x) and delta(p) are standard deviations.
(delta(x))^(2)=|<x^(2)>-<x>^(2)|
(delta(p))^(2)=|<p^(2)>-<p>^(2)|

 

[betel]

One can derive the general uncertainity principle by using the standard deviation method of statistics.
Here delta(x) and delta(p) are standard deviations.
(delta(x))^(2)=|<x^(2)>-<x>^(2)|
(delta(p))^(2)=|<p^(2)>-<p>^(2)|

So how does this help solving the problem at hand?