## Module Question 20

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

Duby3L
Posts: 72
Joined: Fri Sep 28, 2018 12:26 am

### Module Question 20

20. 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 calculated the uncertainty in kinetic energy, but how do we calculate it in energy per moles?

Nicole Lee 4E
Posts: 60
Joined: Fri Sep 28, 2018 12:16 am
Been upvoted: 1 time

### Re: Module Question 20

Multiply your answer by 6.022x10^23 because there are 6.022x10^23 electrons in 1 mol of electrons.

Stevin1H
Posts: 89
Joined: Fri Sep 28, 2018 12:17 am

### Re: Module Question 20

To calculate the energy per mole, you would have to multiply by Avogadro's number. The kinetic energy that you calculate was the kinetic energy for one photon. So, by multiplying by 6.022E23, this will give you the total energy for 1 mole of electrons.

Here is the conversion:

Joules/electron (6.022E23 electrons/1mol electron). The units work out to be Joules/mol.

abbydouglas1K
Posts: 65
Joined: Fri Sep 28, 2018 12:26 am

### Re: Module Question 20

How do we calculate the uncertainty of the kinetic energy ?

melissa_dis4K
Posts: 106
Joined: Fri Sep 28, 2018 12:28 am

### Re: Module Question 20

To calculate the uncertainty in kinetic energy you need to use the uncertainty of velocity from the previous problem: delta v=3.3*10^10 m/s. Then use the formula for kinetic energy= 1/2 Mass * Velocity^2. Plug in the mass of an electron (given) 9.11*10^-31 and the velocity from the previous problem 3.3*10^10 m/s. You will then get that the uncertainty in kinetic energy is 5.3*10^-10 J. To get the uncertainty of kinetic energy per mole you will need to take the found kinetic energy of the electron and multiply it by Avogadro's number 6.022*10^23.