d^(2)psi      -2m[E-U(x)]psi(x)

                     ---------- =    ---------------------              

                      dx^(2)              hbar^(2)

it increases as the particle's kinetic energy decreases

Primary Conditions a Wave Function must obey:

1. psi(x) is a continuous function

2. psi(x)=0 if x is in a region where it is physically impossible for the particle to be.

3. psi(x) --> 0 as x--> +infinity and x --> -infinity

4. psi(x) is a normalized function

1. Determine a pontential-energy function.

2. draw the potential energy curve

3. establish the boundary conditions that the wave function must satisfy.

4. normalize the wave functions

5.draw graphs of psi(x) and lpsi(x)l^(2)

6. determine the allowed energy levels

7. calculate probablities, wavelengths, or other quantities

      deltaE     lEf-Eil

                          f = ------- =  --------

                                  h            h

1. The can move freely between X=0 and X=L at a constant speed; constant K.

2. No matter how much K, the turning points are at 0 and L.

3. The regions x<0 and x>L are forbidden. The particle cannot leave the box.

U(x)= 0            0(</=) X (</=)

U(x)= infinity      X<0  or X>L

1. since U(x)=0 then the second derivative of psi in respect to x is   -2mEpis(x)

                                         =  ------------

                                              hbar^(2)

2. Beta^(2)   2mE

                = ------

                    hbar^(2)

What functions second derivative is a negative constant times the function itself?

psi1(x)=sin(Betax)          psi2(x)=cos(Betax)

2. Then you take the second derivative of each psi and put it in the psi(x)=Apsi1(x) + Bpsi2(x)

Where X=L:

psi(x=L)=Asin(betaL)=0

1. solve for (beta)*(L); so find a constant that helps achieve sin(betaL)=0

2. solve for beta and plug it into psi(x)

since Beta     sqrt(2mEn)    npi

               =   ---------  =  -----       n=1,2,3...

                      hbar           L

2. solve for energy En=n^(2) h^(2)/8mL^(2)=n^(2)E1.

1.these are the only values forE for which there are physically meanigful solutions to the Sch. equation.

2. it is the energy of the stationary state with quantum number n

3. E1 is the ground state energy

1. equation on reference sheet.

2. solve for An, which seems to always equal sqrt(2/L).

a quantum particle in a box cannot be at rest

E1 is the zero-point motion

the fraction of its time that it spends passing through dx

Probclass(in dx at x)=(dt/.5T)= (2/Tv(x))

1. classically forbidden regions are E<U0

2. the wave function on the graph extends into the classically forbidden area

3. there are only a finite number of bound states

4. represent electrons confined to or bound in the potential well

5.no stationary states E>U0

1. because the wave functions are slightly more spread out, you get a lower velocity and thus lower energy.

1. e^(x/P.D.) and e^(-x/P.D.) are used where psi(x)=Ae^(x/P.D.) + Be^(-x/P.D.)   for x greater than or equal to L

2. psi-->0 as x-->infinity

3. since e^(x/P.D.) diverges as x--->infinity A=0

psi(at x=L)= Be^(-L/P.D.)= Psiedge

1. solve for B to get Psi(x)=Psiedge e^-((x-L)/P.D.) for X (>/=)L

P.D.          hbar

      =  ---------------

            sqrt(2m(U0-E))

w=sqrt(k/m)

U(x)=.5kx^(2)

1. A higher total energy line means a larger K not a higer elevation

2. tunnels its way through the hill and emerges on the other side

3. requires not expenditure of energy

En= -13.60eV/n^(2)  n=1,2,3

n is the principal quantum number

L= sqrt(l(l+1)) hbar  l=0,1,2,3..., n-1

l is the orbital quantum number

Lz= (m)(hbar)   m=-l, -l +1,...,0,..., l-1, l

m is the magnetic quantum number

thetalm=cos^(-1)(Lz/L) 

ground state of l=0 has no angular momentum

1. prepare an atomiv beam by evaporating atoms out of a hole in an oven

2. an atom whose magnetic moment vector mnu is tilted upward (mnu >0) has an upward force on it's north pole that is larger than the downward force on its south pole.

3. a downward tilted magnetic moment (mnu<0)experiences a net downward force 

4. A magenetic moment perpendicular to the field (mnu=0) feels no net force and passes through the magenet with no deflection.

1. ms= (+/-)1/2

2. spin up state is the positive and spin down is the negative

3. The electron's spin is an intrinsic quantum property of the electron that has no classical counterpart

delta l= absolute value of l2-l1=1

lamda= (hc)/(deltaEatom)

Nexc=N0e^(-t/T)

where T=1/r

r is the decay rate =Prob(emission in deltat at time t)/(deltat)

Epulse=(Power)(deltat)

Ephoton= hf=(hc)/(lambda)

Epulse=N (Ephoton)

• the atoms of an element with different values of A; where A=Z+N
• only 266 are stable
• isobars- same A but different Z and N

• r=r0A^(1/3) where r0=1.2fm
• all nuclei have the same density of 2.3 x10^(17) kg/m^(3)
• 
  

• The energy you would need to supply to a nucleus to disassemble it into individual protons and neutrons
• B=(ZmH +Nmn - matom) x(931.49)
• peaks ingraph are due to closed shells
• nuclear force is a short-ranged force
• heavier nuclei can become more stable by breaking into smaller pieces
• lighter nuclei can become more stable by fusing together