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CHAPTER 30

RADIOACTIVITY

DISCOVERY – BECQUEREL

DEVELOPMENT – Pierre and Marie Curie

And others

THREE RADIATIONS

ALPHA – nucleus of Helium

BETA - electrons

GAMMA – electromagnetic

2

ALPHA is Nucleus of Helium

BETA PARTICLES are electrons

GAMMA – electromagnetic energy

A High Energy Photon

3

NUCLEAR TRANSMUTATIONS

ALPHA DECAY

�������� �� ℎ� + �ℎ��

�����

���

Masses and Charges must balance

�������� ���

���

��� + �ℎ�����

4

BETA DECAY

�ℎ������� ���� + ����

�����

���

Masses and Charges must balance

�ℎ������� � + ����

������

�����

5

ACTIVITY

NUCLEAR DECAY RATE

NUMBER OF NUCLEAR DECAYS PER

SECOND

ACTIVITY

t

N=

∆

∆

PROPORTIONAL TO NUMBER OF

NUCLEI PRESENT, N

N

t

N∝

∆

∆

6

THEREFORE

N

t

NRATE λ=

∆

∆−=

WHERE λ, THE CONSTANT OF

PROPORTIONALITY, IS THE DECAY

CONSTANT.

Can use calculus to find number of nuclei

(N) as function of time.

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teNN

λ−= 0

START WITH N0 NUCLEI WILL HAVE

AFTER TIME t N.

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HALF LIFE

IF YOU START WITH N0 NUCLEI HOW

MUCH TIME WILL IT TAKE TO ONLY

HAVE HALF THAT NUMBER?

THAT WILL BE CALLED THE HALF-

LIFE.

START WITH N0

WHEN WILL HAVE 2

0N

9

2/1

00

2

teN

N λ−=

SOLVE FOR t1/2

2/1

2

1 te

λ−=

10

λ2

1ln

2/1

−

=t

λ

693.02/1 =t

2/12

1ln tλ−=

11

Insert FIGURE 30.6

EXAMPLE

If a radioactive material initially contains

3.00 mg of U234

92 , how much will remain

unchanged after 62,000 years? (For U234

92

t1/2 = 2.48 x 105 y and λ = 8.88 x 10

-14 s

-1.)

teNN

λ−= 0

( ) atomsxg

atomsxxgxN

1823

3

0 1072.7234

1002.61000.3 == −

12

173.0102.61048.2

693.0 4

5 == yxxyx

tλ

THEREFORE

173.0181072.7 −= eXN

13

atomsXN1810494.6=

MASS = ?

( ) gxatomsx

gxatomsxm

3

23

18 1052.21002.6

23410494.6 −==

WHAT IS THE ACTIVITY OF THIS

SAMPLE?

N

t

NRATE λ=

∆

∆−=

14

( ) ( )atomsXxsxNt

N 18114 10494.61088.8 −−−=−=∆

∆λ

sdisxNt

N/1076.5 5−=−=

∆

∆λ

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DECAY SERIES

HAVE SEEN

������

������� �� ℎ� + �ℎ��

���

And

�ℎ�����

������� ���� + ����

���

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WE CAN FOLLOW A SERIES -

Uranium 238 SERIES

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MOST RADIOACTIVE NUCLEI ARE

IN ONE OF 4 SERIES

�������� ����

�����

���

�ℎ������� ����

�����

���

�������� ����

�� ��

��!

�"������� #$��

�����

���

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NOTE THAT NO SERIES DECAYS TO

�������

IF AT TIME IN HISTORY MOLTEN

LEAD SOLIDIFIED THERE WOULD

BE A MIXTURE OF Pb ISOTOPES.

WHEN IT SOLIDIFIED FOR

EXAMPLE

%&'(

()*

%&'(()+ = -./�0��"�

WITH TIME ������� INCREASES

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AND THUS

�������

�������

123456757

THIS GIVES A METHOD OF

DETERMINING TIME SINCE

SOLIDIFICATION

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ANOTHER AGE DATING METHOD

CARBON 14 DATING

CARBON 12 IS THE MOST

ABUNDENT ISOTOPE OF CARBON

AND IS STABLE

CARBON 14 IS UNSTABLE AND

DECAYS TO NITROGEN (A GAS)

3������� 2 + ����

���

��

WITH HALF-LIFE OF 5730 YEARS

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23

OVER TIME RATIO LIVING PLANTS

AND ANIMALS

8*

9+

8*9( = 3:27�62�;6<�5

FROM TIME OF DEATH ON

3������� 2 + ����

���

��

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AND THEREFORE

8*

9+

8*9( = =�>?��-�-

MEASURE RATIO OF CARBON 14

TO CARBON 12 AND DETERMINE

YEARS SINCE DEATH OF PLANT OR

ANIMAL.

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FISSION

@ + ����������

��! �������� A? + �� + 3@!�

�������

�����

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27

28

FUSION

� + �������� � + C + DE = 0.42J�;�

��� �

�

Then

� + �������� ���

� + KE = 5.49J�;��

��

Then

���� + ��

������� �� + � + ��

� ��

��

��

With E = 12.86J�;

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Can only take place at high temperature

and high pressure.

In Sun

Or Maybe Some Day

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FUNDAMENTAL PARTICLES AND

FORCES

FOUR FORCES FOUND IN NATURE

1. The Strong Interaction

2. The electromagnetic interaction

3. The weak interaction

4. The gravitational interaction

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PARTICLES

Leptons

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Hadrons

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Quarks