I have the following exercise:

Use Heisenberg's uncertainty principle and the relation $\Delta u = \sqrt{\langle u^2 \rangle - \langle u \rangle^2}$ to find the range of energy an electron has in an atom of diameter 1 amstrong.

This is the attempt at a solution:

  • From the uncertainty principle: $\Delta p \Delta x \geq \hslash / 2 $. Therefore $\Delta p \geq \hslash / 2\Delta x$.

  • Without considering relativistic corrections (don't know if this is OK), $E_c = p^2/2m$

  • From the definition of standard deviation $\Delta p = \sqrt{\langle p^2 \rangle - \langle p \rangle^2}$. Then $\Delta p^2 = \langle p^2 \rangle - \langle p \rangle^2$. Therefore $\langle p^2 \rangle = \Delta p^2 + \langle p \rangle^2$

  • The energy of the electron will be the kinetic energy minus the potential energy:

$E = p^2/2m - e^2/r$

And so the average energy will be

$\langle E \rangle = \langle p^2\rangle/2m - e^2/\langle r \rangle = (\Delta p^2 + \langle p \rangle^2)/2m - e^2/\langle r \rangle$

Here I don't know how to continue. Do I have to assume $\langle p \rangle = 0$? Why? And even when I assume that, replacing $\langle r\rangle$ with $\Delta x$, which I suppose is $1 * 10^{-10} m$ (why?) I don't get a value of energy similar to the ground level energy of a hydrogen atom (which has roughly the same diameter than this one).

What am I doing wrong?


1 Answer 1


$\langle {\bf p}\rangle=0$ because momentum is a vector and it points in all directions with equal probability assuming the wavefunction $\psi( {\bf r})$ is spherically symmetrical.

$\langle {\bf |r|}\rangle \approx a_0$.

Putting these together gives the lower bound

$\langle E \rangle = \langle {\bf p}^2 \rangle/2m - e^2/4 \pi \epsilon _0 \langle | {\bf r} | \rangle =\langle {\bf \Delta p} \rangle ^2 /2m - e^2/4 \pi \epsilon _0 \langle | {\bf r} | \rangle \approx \hbar ^2/8ma_0-e^2/4 \pi \epsilon _0 a_0 \approx-24 eV $

which is pretty close to the actual binding energy $-13.6 eV$.

  • $\begingroup$ Thank you. You are right, that seems to be what the problem was aiming at. I thought that the problem still considered Bohr's atom model with orbits, therefore <p> couldn't be 0. $\endgroup$
    – Heathcliff
    Apr 20, 2013 at 18:14

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