**In an experiment, with protons moving through a potential difference V = 4,000 volts, the kinetic energy gained should be equal to the work done, or:**

*(1)***½**

**m**

_{p}v^{2 }= eV

Where

**m**the proton mass, is:_{p }*1.7 x 10*^{-27}kg
And
the proton charge = e = 1.6 x 10

^{-19}C
The
velocity then is:

v
= Ö (2eV/m

**)**_{p}**v = m**

_{p}**Ö (2eV/m**

_{p}**) = Ö (2eV m**

_{p}**)**

_{p}
l

_{D}= h/p = h/Ö (2eVm**)**_{p}
For
a voltage V = 4,000 V one would find:

l

_{D}=
(6.626
x 10

^{-34}J-s) / [(3.2 x 10^{-19}C) (4000V) ( 1.7 x 10^{-27}kg)]^{1/2}

l

_{D}= 4.49 x 10^{-13}m

**Write out the electron configuration for oxygen (O16), then write out the values for the set of quantum numbers : n, ℓ, m**

*(2)*_{ℓ}, m

_{s}for each of the electrons in O16

Finding the electron configuration means one must use the

**Pauli Exclusion Principle**to make sure the electrons are distributed so that

**. This means we need to have 2 in the 1s shell, 2 in the 2s shell, and 4 in the 2p shell, so:**

*no two electrons have the same set of quantum numbers*1s(2) 2s(2) 2p(4)

For the 2 electrons in the 1s shell: n = 1, ℓ = 0, = 0 and m

_{s}= ½ (and (- ½), for second)

For the 2 electrons in the 2s shell: n=2, ℓ =0, m

_{ℓ}= 0 and m

_{s}= ½ (and (- ½), for second)

For the 4 electrons in the 2p shell:

Since ℓ =1 corresponds to p: n=2, ℓ =1, m

_{ℓ}= +1, m

_{s}= + ½

Then: n=2, ℓ =1, m

_{ℓ}= -1, m

_{s}= -½

And: n=2, ℓ =1, m

_{ℓ}= 0, m

_{s}= +½

Finally: n=2, ℓ =1, m

_{ℓ}= 0, m

_{s}= -½

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