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Chapter 2

2.1 Sketch_______________________________________

2.2 Sketch_______________________________________

2.3 Sketch_______________________________________

2.4

From Problem 2.2, phase

= constant Then

From Problem 2.3, phase

= constant Then

_______________________________________

2.5

Gold: eV J

So,

cm or m Cesium: eV

J So,

cm or m_______________________________________

2.6

(a)

kg-m/s

m/s

or cm/s

(b)

kg-m/s

m/s

or cm/s (c) Yes_______________________________________

2.7(a) (i)

kg-m/s

m

or

(ii)

kg-m/s


m

or

(iii)

kg-m/s

m

or

(b)

kg-m/s

m

or

_______________________________________

2.8

eV Now

or kg-m/s Now

m

or

_______________________________________

2.9

Now

and

Set and Then

which yields

J keV_______________________________________

2.10

(a)

kg-m/s

m/s

or cm/s

J

or eV

(b)

J

or eV


kg-m/s

m

or

_______________________________________

2.11(a)

J Now

V kV(b)

kg-m/s Then

m

or

_______________________________________

2.12

kg-m/s_______________________________________

2.13(a) (i)

kg-

m/s

(ii)

Now

kg-m/sso

J

or eV

(b) (i) kg-m/s (ii)

kg-m/s

J

or

eV_______________________________________

2.14

kg-m/s

m/s_______________________________________

2.15(a)

s


(b)

kg-m/s_______________________________________

2.16(a) If and are solutions

to Schrodinger's wave equation, then

and

Adding the two equations, we obtain

which is Schrodinger's wave equation. So is also a solution.

(b) If were a solution toSchrodinger's wave equation, then we could write

which can be written as

Dividing by , we find

Since is a solution, then

Subtracting these last two equations, we have

Since is also a solution, we have

Subtracting these last two equations, we obtain

This equation is not necessarily valid, which means that is, in general, not a solution to Schrodinger's wave equation._______________________________________

2.17

so

or

_______________________________________

2.18


or _______________________________________

2.19

Note that

Function has been normalized.(a) Now

or

which yields (b)

or

which yields (c)

which yields

_______________________________________

2.20

(a)

or

(b)


or

(c)

or _______________________________________

2.21

(a)

or

(b)

or

(c)

or _______________________________________

2.22

(a) (i) m/s

or cm/s

m

or

(ii) kg-m/s

J

or eV

(b) (i) m/s

or cm/s

m

or

(ii) kg-m/s eV

_______________________________________

2.23(a)


(b)

so m/scm/sFor electron traveling in direction, cm/s

kg-m/s

m

m

or rad/s_______________________________________

2.24(a) kg-m/s

m

m

rad/s(b) kg-m/s

m

m

rad/s

_______________________________________

2.25

J or

or eV Then eV

eV

eV_______________________________________

2.26(a)

Jor

eV

Then eV eV eV

(b)

J

mor nm

_______________________________________

2.27


(a)

or (b) mJ(c) No

_______________________________________

2.28 For a neutron and :

J or

eV

For an electron in the same potential well:

J or

eV

_______________________________________

2.29 Schrodinger's time-independent wave equation

We know that

for and

We have

for

so in this region

The solution is of the form

where

Boundary conditions:

at

First mode solution: where

Second mode solution: where

Third mode solution: where

Fourth mode solution: where

_______________________________________

2.30 The 3-D time-independent wave equation in cartesian coordinates for is:

Use separation of variables, so let Substituting into the wave equation, we obtain


Dividing by and letting ,

we find

(1)

We may set

Solution is of the form

Boundary conditions:

and

where Similarly, let

and

Applying the boundary conditions, we find

,

,

From Equation (1) above, we have or

so that

_______________________________________

2.31 (a)

Solution is of the form:

We find

Substituting into the original equation, we find:

(1)

From the boundary conditions,

, where

So ,

Also , where

So ,

Substituting into Eq. (1) above

(b)Energy is quantized - similar to 1-D result. There can be more than one quantum state per given energy - different than 1-D result.

_______________________________________

2.32(a) Derivation of energy levels exactly the

same as in the text

(b)

For Then

(i) For

J


or

eV

(ii) For cm

J or

eV

_______________________________________

2.33(a) For region II,

General form of the solution is

where

Term with represents incident wave and term with represents reflected wave. Region I,

General form of the solution is

where

Term involving represents the transmitted wave and the term involving represents reflected wave: but if a particle is

transmitted into region I, it will not be reflected so that . Then

(b) Boundary conditions:(1)

(2)

Applying the boundary conditions to the solutions, we find

Combining these two equations, we find

The reflection coefficient is

The transmission coefficient is

_______________________________________

2.34

where

m

(a) For m


(b) For m

(c) For m

_______________________________________

2.35

where

or m(a) For m

(b) For m

(c) , where is the density of transmitted electrons. eV J

m/scm/s

electrons/cm

Density of incident electrons,

cm

_______________________________________

2.36

(a) For

or m Then

or

(b) For

=

or m Then

or _______________________________________


2.37

where

m (a)

(b)

or m_______________________________________

2.38 Region I , ; Region II , Region III ,

(a) Region I:

(incident) (reflected)

where

Region II:

where

Region III:

(b) In Region III, the term represents a reflected wave. However, once a particle is transmitted into Region III, there will not be a reflected wave so that . (c) Boundary conditions: At :

At :

The transmission coefficient is defined as

so from the boundary conditions, we want to solve for in terms of . Solving for in terms of , we find

We then find


We have

If we assume that , then will be large so that We can then write

which becomes

Substituting the expressions for and , we find

and

Then

Finally,

_____________________________________

2.39 Region I:

incident reflected where

Region II:

transmitted reflected where

Region III:

transmitted where


There is no reflected wave in Region III. The transmission coefficient is defined as:

From the boundary conditions, solve for in terms of . The boundary conditions are: At :

At :

But

Then, eliminating , , from the boundary condition equations, we find

_______________________________________

2.40(a) Region I: Since , we can write

Region II: , so

Region III: The general solutions can be written, keeping in mind that must remain finite for , as

where

and

(b) Boundary conditions At :

At :

or (c)

and since , then

From , we can write

or

This equation can be written as


or

This last equation is valid only for specific values of the total energy . The energy levels are quantized.

_______________________________________

2.41

(J)

(eV)

or

(eV)

eV eV eV eV_______________________________________

2.42 We have

and

or

To find the maximum probability

which gives

or is the radius that gives the greatest probability._______________________________________

2.43 is independent of and , so the wave equation in spherical coordinates reduces to

where

For

Then

so

We then obtain

Substituting into the wave equation, we have


where

Then the above equation becomes

or

which gives 0 = 0 and shows that is indeed a solution to the wave equation._______________________________________

2.44 All elements are from the Group I column of the periodic table. All have one valence electron in the outer shell._______________________________________

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