© J Parkinson 1 2 Mass Defect The difference between the mass of the atom and the sum of the masses...

Post on 20-Jan-2016

214 views 0 download

Tags:

Transcript of © J Parkinson 1 2 Mass Defect The difference between the mass of the atom and the sum of the masses...

© J Parkinson1

2

Mass  Defect

The difference between the mass

of the atom and the sum of the masses

of its parts is called  the  mass defect  (m).    

Careful measurements show that the mass of a particular atom is always slightly less than the sum of the masses of the individual protons, neutrons and electrons of which the atom consists.

e.g. a helium nucleus consists of 2 protons and 2 neutrons

2 protons & 2 neutrons Helium atom

3

mp = mass of a proton   (1.007277 amu)

mn = mass of a neutron (1.008665 amu)

me = mass of an electron   (0.000548597 amu)

1 atomic mass unit ( amu ) = 1.661 x 10-27 kg

m = [ Z(mp + me) + (A-Z)mn ] - matom

1 amu = 1/12 of the mass of an atom of Carbon-12

Z = PROTON NUMBER (ATOMIC NUMBER) AND ALSO EQUALS THE NUMBER OF ELECTRONS

A = NUCLEON NUMBER (MASS NUMBER) AND ALSO EQUALS THE NUMBER OF

PROTONS + NEUTRONS

THE NUMBER OF NEUTRONS MUST BE A - Z

4

For Helium the mass defect will be

m = [ 2(1.007277 + 0.000548597 ) + 2 x 1.008665 ] – 4.021435 = 0. 0115462 amu

0. 0115462 amu x 1.661 x 10-27 = 1. 91782 x 10-29 kg

Using Einstein’s E = mc2, this is equivalent to a loss of energy given by

E = 1.91782 x 10-29 x (3 x 108)2 Joules = 1.726 x 10-12 J

This figure is the BINDING ENERGY of the Helium nucleus.

THE BINDING ENERGY of a nucleus is defined as the energy which must be input to separate all of its protons and neutrons.

5

Binding Energies are usually expressed in MeV 1 amu = 931.3 MeV

The binding energy of the Helium nucleus is therefore

0. 0115462 amu x 931.3 = 10.75 MeV

To compare the stabilities of different nuclei,

Binding Energies PER NUCLEON in the nucleus are compared.

For Helium this will be 10.75 ÷ 4 = 2.69 MeV per nucleon

The higher the binding energy per nucleon, the greater the stability of the nucleus

6

NUCLEON NUMBER

BINDING ENERGY

Per NUCLEON

2H

238U

56Fe8.8 MeV

Iron is the most stable nucleus

7

BE/A

A

ALTERNATIVELYBE/A

A0 250

9.0

As energy is lost when nuclei are synthesised

8

FISSION

IRON

Heavy nuclei may increase their stability by Nuclear Fission

Light nuclei may increase their stability by Nuclear Fusion

FUSION

9

As atomic mass increases, the neutron to proton ratio for stable nuclei increases because proton-proton repulsion becomes significant, as the number of protons increases. Cohesive nuclear forces arise form neutrons, so the neutron to proton ratio must increase for heavier elements.

Belt of StabilityBelt of Stability

Proton number, Z

Ne

utr

on

nu

mb

er,

N =

A -

Z

N = Z

Bel

t of S

tabl

e Is

otop

es

For helium He- 4 the N:P ratio is 1 : 1

For uranium U- 238 the ratio is 1 : 1.6

10

Nuclear Fission is the fragmentation of heavy nuclei to form lighter, more stable ones.

The Fission of U - 235

U23592

energyofMeVnKrBanU 1803 10

9436

13956

10

23592

This is only one of several fissures that are possible.

On average 2.5 neutrons are released

© J Parkinson11

Neutrons released in the fission of 235U can induce three more fissions, then nine, and so on leading to a chain reaction.

12

Critical mass is the mass required for the chain reaction to become self-sustaining.

neutron

Some neutrons may :

Cause more fission

Get lost

Be absorbed by an atom

lost

For a chain reaction to be self sustaining, every fission must produce at least one more

neutron that will initiate further fissions.

13

Nuclear Reactor

Pressure vessel and biological shield

Graphite Moderator slows neutrons to thermal energies [1 MeV]

Fuel Rods remain in reactor until spent

Coolant [CO2 or H2O] to extract energy

To heat exchanger where steam is produced

Boron control rods absorb neutrons to control the reaction