Physics 1C Lecture 29B

"Nuclear powered vacuum cleaners will probably be a reality within 10 years. "

--Alex Lewyt, 1955

Problems with Bohr’s Model Bohr’s explanation of atomic spectra includes some

features of the currently accepted theory.

Bohr’s model includes both classical and non-classical ideas.

He applied Planck’s ideas of quantized energy levels to orbiting electrons and Einstein’s concept of the photon to determine frequencies of transitions.

Improved spectroscopic techniques, however, showed that many of the single spectral lines were actually groups of closely spaced lines.

Single spectral lines could be split into three closely spaced lines when the atom was placed in a magnetic field.

Quantum Numbers n is considered the principal quantum number.

n can range from 1 to infinity in integer steps.

But other quantum numbers were added later on to explain other subatomic effects (such as elliptical orbits; Arnold Sommerfeld (1868-1951).

The orbital quantum number, ℓ, was then introduced. ℓ can range from 0 to n-1 in integer steps.

Historically, all states with the same principal quantum number n are said to form a shell. Shells are identified by letters: K, L, M, …

All states with given values of n and ℓ are said to form a subshell. Subshells: s, p, d, f, g, h, …

Quantum

Numbers Largest angular

momentum will be

associated with:

ℓ = n – 1

This corresponds to

a circular orbit.

ℓ = 0 corresponds

to spherical

symmetry with no

axis of rotation.

Quantum Numbers

Suppose a weak magnetic field is applied to an atom and its direction coincides with z axis.

Then direction of the angular momentum vector relative to the z axis is quantized!

Lz = mℓ ħ Note that L cannot be

aligned with the z axis.

Because angular momentum is a vector, its direction

also must be specified.

Quantum Numbers States with quantum numbers

that violate the rules below cannot

exist.

They would not satisfy the

boundary conditions on the wave

function of the system.

Quantum Numbers In addition, it becomes

convenient to think of the electron as spinning as it orbits the nucleus.

There are two directions for this spin (up and down).

Another quantum number accounts for this, it is called the spin magnetic quantum number, ms.

Spin up, ms = 1/2

Spin down, ms = –1/2.

Wave Functions for Hydrogen

13

1( ) or as

o

r ea

The simplest wave function for hydrogen is the one

that describes the 1s state:

All s states are spherically symmetric.

The probability density for the 1s state is:

2 2

1 3

1or a

s

o

ea

Wave Functions for Hydrogen The radial probability density for the 1s state:

The peak at the Bohr radius indicates the most

probable location.

22

1 3

4( ) or as

o

rP r e

a

The atom has

no sharply

defined boundary.

The electron

charge is

extended through

an electron cloud.

Wave Functions for Hydrogen The electron cloud model is quite different from the

Bohr model.

The electron cloud structure does not change with time and remains the same on average.

The atom does not radiate when it is in one particular quantum state.

This removes

the problem of the

Rutherford atom.

Radiative

transition causes

the structure to

change in time.

Wave Functions for Hydrogen The next simplest wave function for hydrogen is for

the 2s state n = 2; ℓ = 0.

This radial probability

density for the 2s state

has two peaks.

In this case, the most

probable value is r 5a0.

The figure compares 1s

and 2s states. 3

22

2

1 1( ) 2

4 2

or a

s

o o

rr e

a a

Clicker Question 29B-1

Which of the following hydrogen atom series can

emit a photon with a wavelength in the infrared part

of the spectrum?

A) Lyman

B) Balmer

C) Paschen

D) All of the above series can emit a photon

with a wavelength in the infrared region

Problem

If an electron has a measured deBroglie wavelength

of 0.850x10–10m, what is its kinetic energy?

a. 55.0 eV.

b. 104 eV.

c. 147 eV.

d. 207 eV.

e. 18.8 eV.

Solution If an electron has a measured deBroglie wavelength of

0.850x10–10m, what is its kinetic energy?

What do we know?

deBroglie wavelength ->

Momentum – l = h/p

What do we want?

KE. Which is related to momentum

K.E. = (1/2)p2/m = (1/2)(h/l)2/m

= (1/2)(6.626x10-34Js/0.85x10-10m)2/(9.11x10-31kg)

x(eV/1.602x10-19J)

= 207 eV (choice d)

Electron Paramagnetic

Resonance (EPR) Apply magnetic field to

remove energy

degeneracy of spin up

and spin down electrons

When “splitting” matches

photon energy, get

absorption

Seen as a signal

Relatively rare in nature

The Nucleus All nuclei are composed of protons and neutrons

(they can also be called nucleons).

The atomic number, Z, is the number of protons in

the nucleus.

The neutron number, N, is the number of neutrons

in the nucleus.

The mass number, A, is the number of nucleons in

the nucleus (A = N + Z).

In symbol form, it is: XAZ where X is the chemical symbol of the element.

The Nucleus Isotopes of an element have the same Z but

differing N (and thus A) values.

For example,

both have 92 protons, but U-235 has 143 neutrons

while U-238 has 146 neutrons.

Nucleons and electrons are very, very small. For

this reason, it is convenient to define a new unit of

mass known as the unified mass unit, u.

Where: 1 u = 1.660559x10-27 kg (exactly 1/12 of

the mass of one atom of the isotope C-12)

92

235U

92

238U

The Nucleus Einstein’s mass-energy equivalence tells us that all

mass has an energy equivalence. This is known as

the rest energy.

This means mass can be converted to some

form(s) of useable energy.

The equation is given by: From here we get that the energy equivalent of 1u

of mass is:

ER mc2

ER 1 u c2 1.4924311010 J 931.494 MeV

This gives us the relationship between mass and

energy to also be:

1 u 931.494 MeVc2

Nuclear Stability Example

What is the binding energy (in MeV) of the

Helium-4 nucleus? The atomic mass of

Helium is 4.002602u.

Answer

We need to calculate the total rest energy of the

Helium nucleus and the total rest energy of the

separated nucleons.

We need to turn to: ER = mc2

Nuclear Stability Answer

For the Helium nucleus we find:

For the separated protons and neutrons:

ER mHec2 4.002602 u c2

ER 4.002602 u c2 931.494

MeVc2

1 u 3,728.4 MeV

ER 2mpc2 2mnc

2

ER 2 938.28 MeV

c2 c2 2 939.57 MeV

c2 c2 3,755.7 MeV

Calculate the difference:

Eb EHe Esep 3,728.4 MeV 3,755.7 MeV27.3 MeV

Ebnucleon

27.3 MeV4 6.825 MeVA

The Nucleus This table gives the following masses for the

selected particles:

Note that the neutron is slightly heavier than the

proton.

Nuclear Magnetic

Resonance Imaging Nuclei also have quantum

numbers

Detect nuclear magnetic

moments similar to NMR

Nuclear Magnetic

Resonance Imaging

magnetic moments of some of these protons change

and align with the direction of the field

radio frequency transmitter is briefly turned on,

producing a further varying electromagnetic field

photons of this field have just the right energy, known

as the resonance frequency, to be absorbed and flip

the spin of the aligned protons

Nuclear Magnetic

Resonance Angiography

generates pictures of

the arteries

administration of a

paramagnetic contrast

agent (gadolinium) or

using a technique

known as "flow-related

enhancement"

For Next Time (FNT)

Finish the homework for Chapter 29

Read Chapter 30 through page

1022

Wednesday – Overview of quarter