Solutions to Quantum Mechanics Problems - Pt. 1

(1)  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:

½ m _p v²   =   eV

 

Where m _p  the  proton mass,  is: 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)

 

The momentum p = m _pv = m_p Ö (2eV/m_p) = Ö (2eV m _p)

 

And the de Broglie wavelength would be:

 

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

 

(2)  Write out the electron configuration for oxygen (O16), then write out the values for the set of quantum numbers : n, ℓ, m _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 no two electrons have the same set of quantum numbers. This means we need to have 2 in the 1s shell, 2 in the 2s shell, and 4 in the 2p shell, so:

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