## Post Module Question #8/9 [ENDORSED]

$\Delta p \Delta x\geq \frac{h}{4\pi }$

Kendall_Chaffin_3C
Posts: 25
Joined: Wed Sep 21, 2016 2:59 pm

### Post Module Question #8/9

Question 8 states If one incorrectly assumed that an electron is located inside the nucleus of an atom, then for a hydrogen atom the electron is confined to its nuclear diameter of 1.7 x 10-15 m which would be the electron's uncertainty in position. Use the Heisenberg uncertainty equation to calculate the electron's uncertainty in momentum. Then use the mass of electron (9.1 x 10-31 kg) to calculate its uncertainty in velocity.
Comment on your value.

I got that Delta p <= 3.1 x 10-20kg.m.s-1, Delta v = 3.4 x 1010m.s-1

For question 9, it says use the above uncertainty in velocity to calculate the electron's uncertainty in kinetic energy. Then calculate the uncertainty in kinetic energy per mole of electrons (that is, per mole of hydrogen atoms). Comment on your value.

I used the velocity I got to determine the wavelength then used the wavelength to determine the energy; but, my answer didn't match any of the options. What am I doing wrong?

kara_kremer_2N
Posts: 20
Joined: Fri Jul 15, 2016 3:00 am

### Re: Post Module Question #8/9  [ENDORSED]

Try using the equation KE=1/2 mv^2

You should get the kinetic energy for one hydrogen atom. From there you can use dimensional analysis to calculate the kinetic energy per mole of electrons/hydrogen atoms.

FizaBaloch1J
Posts: 41
Joined: Fri Apr 06, 2018 11:01 am

### Re: Post Module Question #8/9

kara_kremer_2N wrote:Try using the equation KE=1/2 mv^2

You should get the kinetic energy for one hydrogen atom. From there you can use dimensional analysis to calculate the kinetic energy per mole of electrons/hydrogen atoms.

How do we "calculate the kinetic energy per mole of electrons/hydrogen atoms" because once we find the KE, isn't that J per electron, so then do we divide our answer in J by avogadros number to cancel out electrons and get moles?

NatalieSDis1A
Posts: 39
Joined: Fri Apr 06, 2018 11:05 am

### Re: Post Module Question #8/9

I think the key here is realizing that Hydrogen has 1 electron so if you have a certain amount of electrons you have that amount of hydrogen atoms. I believe thats the why the hydrogen add on is in parentheses, signifying that the answer is the same for j/mol of electron and j/ mol of hydrogen atom.

The conversion should be something like this:
(X joules/ 1 electron)(6.02 x 10^23 electrons/ 1 mol electrons)(1 mol electrons/1 mol of hydrogen atoms)