Consider the nuclear reaction: 11 Na 23 + α → [ 13 Al 27 ] → 12 Mg 26 + 1 H 1 Z X A + α...

22
ARTIFICIAL RADIOACTIVITY

Transcript of Consider the nuclear reaction: 11 Na 23 + α → [ 13 Al 27 ] → 12 Mg 26 + 1 H 1 Z X A + α...

Page 1: Consider the nuclear reaction:   11 Na 23 + α → [ 13 Al 27 ] → 12 Mg 26 + 1 H 1   Z X A + α → [ Z+2 C n A+4 ] → Z+1 Y A+3 + 1 H 1  Target Compound.

ARTIFICIAL RADIOACTIVITY

Page 2: Consider the nuclear reaction:   11 Na 23 + α → [ 13 Al 27 ] → 12 Mg 26 + 1 H 1   Z X A + α → [ Z+2 C n A+4 ] → Z+1 Y A+3 + 1 H 1  Target Compound.

ARTIFICIAL RADIOACTIVITY AND Q-VALUE

Consider the nuclear reaction: 

11Na23 + α → [13Al27] → 12Mg26 + 1H1

  ZXA + α → [Z+2Cn

A+4] → Z+1YA+3 + 1H1

Target Compound Product nuclei

 The above reaction is an example for (α, P) reactions.

Page 3: Consider the nuclear reaction:   11 Na 23 + α → [ 13 Al 27 ] → 12 Mg 26 + 1 H 1   Z X A + α → [ Z+2 C n A+4 ] → Z+1 Y A+3 + 1 H 1  Target Compound.

Consider a nuclear reaction represented by the equation:x + X → Y + y

Bombarding Target Product Product

particle nucleus nucleus particleAssume that the target nucleus is at rest and has no K.E.

The total energy of a particle atom = R.E + K.E. Rest energy

Page 4: Consider the nuclear reaction:   11 Na 23 + α → [ 13 Al 27 ] → 12 Mg 26 + 1 H 1   Z X A + α → [ Z+2 C n A+4 ] → Z+1 Y A+3 + 1 H 1  Target Compound.

( Ex + mxC2 + )MXC2 = (EY + MYC2) + (Ey + myC2) (1)

K.E R.E R.E K.E R.E K.E R.E

The difference between K.E of the products of the reaction and that of the incident particle is called the energy balance of the reaction or Q-value.

  Q = EY + Ey – Ex

 In terms of the masses:

EY + Ey – Ex = (MX +mx – MY – my) C2

 From equation (1):

Q = EY + Ey – Ex = (MX + mx – MY – my) C2

Page 5: Consider the nuclear reaction:   11 Na 23 + α → [ 13 Al 27 ] → 12 Mg 26 + 1 H 1   Z X A + α → [ Z+2 C n A+4 ] → Z+1 Y A+3 + 1 H 1  Target Compound.

If the value of Q is +ve, the K.E of the products is greater than that of the reactant, and the reaction is said to be exothermic or exogeric.

 If Q is -ve, the reaction is endothermic or endoergic. The Q-value is one of the main sources of

information about nuclear properties.From the Q-value and the known atomic mass of B11,

He4, and N14, the mass of the neutron can be obtained (Chadwick).

Page 6: Consider the nuclear reaction:   11 Na 23 + α → [ 13 Al 27 ] → 12 Mg 26 + 1 H 1   Z X A + α → [ Z+2 C n A+4 ] → Z+1 Y A+3 + 1 H 1  Target Compound.

THE TRANSURANIUM ELEMENTS

Artificial radioactivity can be achieved not only with isotopes of elements but also with transuranium elements.

  These are elements of atomic no. greater than 92. These elements have been generated by bombarding U (238) with

neutrons or α-particles.       EX: 92U238 + 0n1 → (92U239)* + γ   (92U239)* → 93Np239 + -1β0 ; Np: Neobium.   (93Np239)* → 94Pu239 + -1β0 ; Pu: Plutonium.

Page 7: Consider the nuclear reaction:   11 Na 23 + α → [ 13 Al 27 ] → 12 Mg 26 + 1 H 1   Z X A + α → [ Z+2 C n A+4 ] → Z+1 Y A+3 + 1 H 1  Target Compound.

All of the transuranium elements are radioactive, all are α-emitters and some both α-emitters and β-emitters.

The most important of these elements from the point of view of nuclear engineering is plutonium (239) because of its usefulness as a nuclear fuel.

Page 8: Consider the nuclear reaction:   11 Na 23 + α → [ 13 Al 27 ] → 12 Mg 26 + 1 H 1   Z X A + α → [ Z+2 C n A+4 ] → Z+1 Y A+3 + 1 H 1  Target Compound.

ALPHA DECAY

Page 9: Consider the nuclear reaction:   11 Na 23 + α → [ 13 Al 27 ] → 12 Mg 26 + 1 H 1   Z X A + α → [ Z+2 C n A+4 ] → Z+1 Y A+3 + 1 H 1  Target Compound.

1- THE VELOCITY AND ENERGY OF ALPHA PARTICLES:

The method that gives the velocity and energy of α-particles depends on the measurement of the defection of the paths of the particles in a magnetic field.

  When a charged particle moves in a magnetic field, its

orbit is a circle whose radius is determined from the relation:

qBv = Mv2 / r   Also, v = qBr / M

Page 10: Consider the nuclear reaction:   11 Na 23 + α → [ 13 Al 27 ] → 12 Mg 26 + 1 H 1   Z X A + α → [ Z+2 C n A+4 ] → Z+1 Y A+3 + 1 H 1  Target Compound.

The velocity can be determined if the strength of the magnetic field is known and if the radius of the orbit is measured.

Also, the k.e = 1/2 Mv2

Page 11: Consider the nuclear reaction:   11 Na 23 + α → [ 13 Al 27 ] → 12 Mg 26 + 1 H 1   Z X A + α → [ Z+2 C n A+4 ] → Z+1 Y A+3 + 1 H 1  Target Compound.

2 -NUCLEAR ENERGY LEVELS (DISINTEGRATION ENERGY):

The total energy change in an α-decay process is called α-disintegration energy.

  When an α-particle is emitted, the product, or residual nucleus, carrying with it a certain amount of energy.

The α-disintegration energy is the sum of kinetic energies of the α-particle and the product nucleus, and is found as follows:

From the low of conservation of momentum:  MV = MrVr

  Mass of Its Mass of Product

α-particle velocity product velocity

Page 12: Consider the nuclear reaction:   11 Na 23 + α → [ 13 Al 27 ] → 12 Mg 26 + 1 H 1   Z X A + α → [ Z+2 C n A+4 ] → Z+1 Y A+3 + 1 H 1  Target Compound.

  →Vr = (M / Mr) V → (1)

 The α-disintegration energy is:

Eα = 1/2 MV2 + 1/2 MrVr2

 Substitute from equation (1):

  Eα = 1/2 MV2 + 1/2 Mr [(M2/Mr

2) V2]

  Eα = 1/2 MV2 [1 + (M/Mr)]

   

Page 13: Consider the nuclear reaction:   11 Na 23 + α → [ 13 Al 27 ] → 12 Mg 26 + 1 H 1   Z X A + α → [ Z+2 C n A+4 ] → Z+1 Y A+3 + 1 H 1  Target Compound.

INTERACTION OF RADIATION WITH MATTER

Page 14: Consider the nuclear reaction:   11 Na 23 + α → [ 13 Al 27 ] → 12 Mg 26 + 1 H 1   Z X A + α → [ Z+2 C n A+4 ] → Z+1 Y A+3 + 1 H 1  Target Compound.

The interaction of radiation with matter depends on:i) The type and energy of the incident radiation.ii) The chemical and physical properties of the target

material.iii) The manner in which the incident radiation interacts with

the material. This section contains the mechanisms by which ionizing

radiation interacts and loses energy as it moves through matter. This subject is extremely important for radiation measurements because the detection of radiation is based on its interactions and the energy deposited in the material of which the detector is made. Therefore, to be able to build detectors and interpret the results of the measurement, we need to know how radiation interacts and what the consequences are of the various interactions.

Page 15: Consider the nuclear reaction:   11 Na 23 + α → [ 13 Al 27 ] → 12 Mg 26 + 1 H 1   Z X A + α → [ Z+2 C n A+4 ] → Z+1 Y A+3 + 1 H 1  Target Compound.

For the discussion that follows, ionizing radiation is divided into three groups:

Charged particles: electrons (e-), positrons (e+), protons (p), deutero- ns (d) , alphas (α), heavy ions (A > 4).

Photons: gammas (γ) or X-rays. Non-charged particles: Neutrons (n). This classification is convenient because

each group has its own characteristic properties and can be studied separately.

Page 16: Consider the nuclear reaction:   11 Na 23 + α → [ 13 Al 27 ] → 12 Mg 26 + 1 H 1   Z X A + α → [ Z+2 C n A+4 ] → Z+1 Y A+3 + 1 H 1  Target Compound.
Page 17: Consider the nuclear reaction:   11 Na 23 + α → [ 13 Al 27 ] → 12 Mg 26 + 1 H 1   Z X A + α → [ Z+2 C n A+4 ] → Z+1 Y A+3 + 1 H 1  Target Compound.

Ionizing Radiation

Charged Particles

Heavy Particles P, α, heavy ions

Light Particlese- , e+

Photons (X-ray, γ-ray) Neutrons

Page 18: Consider the nuclear reaction:   11 Na 23 + α → [ 13 Al 27 ] → 12 Mg 26 + 1 H 1   Z X A + α → [ Z+2 C n A+4 ] → Z+1 Y A+3 + 1 H 1  Target Compound.

MECHANISMS OF CHARGED-PARTICLE ENERGY LOSS

Heavy charged particles, such as alpha particle, interact with matter primarily through coulomb forces between their positive charge and the negative charge of the orbital electrons of the absorbed atoms. Although interactions of the particle with nuclei as in Rutherford scattering or alpha particle induced reactions are also possible, such encounters occur only rarely and they are not normally significant in the response of radiation detectors. Instead, charged particle detectors must rely on the results of interactions with electrons for their response.

Page 19: Consider the nuclear reaction:   11 Na 23 + α → [ 13 Al 27 ] → 12 Mg 26 + 1 H 1   Z X A + α → [ Z+2 C n A+4 ] → Z+1 Y A+3 + 1 H 1  Target Compound.

Upon entering any absorbing medium, the charged particle immediately interacts simultaneously with many electrons. The electron feels an impulse from the attractive coulomb force as the particle passes its vicinity. This impulse may be sufficient either to raise the electron to a higher-lying shell within the absorber atom (excitation) or to remove the electron from the atom (ionization).

 

Page 20: Consider the nuclear reaction:   11 Na 23 + α → [ 13 Al 27 ] → 12 Mg 26 + 1 H 1   Z X A + α → [ Z+2 C n A+4 ] → Z+1 Y A+3 + 1 H 1  Target Compound.

On the other hand, when energetic electrons penetrate materials, they lose energy by two mechanisms:

Collision loss, where energy is given to electrons in the atoms of the material, and radiation loss involving the conversion of electron kinetic energy to photons of X-radiation in the field of an atomic nucleus. As the incident electron traveling through the material, it might pass by a particular atom in about 10-18 s so it is able to exert large coulomb forces on the atomic electrons and impart energy to them. The energy transfer may be sufficient to allow the electron to leave the parent atom, and so cause ionization, which is completed within about 10-15 s. Alternatively, the atomic electron may be excited to a higher state. These inelastic collisions are the most important mechanisms of energy loss.

Page 21: Consider the nuclear reaction:   11 Na 23 + α → [ 13 Al 27 ] → 12 Mg 26 + 1 H 1   Z X A + α → [ Z+2 C n A+4 ] → Z+1 Y A+3 + 1 H 1  Target Compound.

Emission of radiation is a mechanism for electron energy loss at high kinetic energies where the electron behavior is relativistic. An electron in the electrostatic field of an atomic nucleus can experience a large acceleration. The rate of energy loss due to radiation increases with the atomic number of the absorbing material and the kinetic energy of the electron. This energy is lost in the form of a photon of X-radiation, often referred to as bremsstrahlung, or braking radiation

Page 22: Consider the nuclear reaction:   11 Na 23 + α → [ 13 Al 27 ] → 12 Mg 26 + 1 H 1   Z X A + α → [ Z+2 C n A+4 ] → Z+1 Y A+3 + 1 H 1  Target Compound.

IN OTHER WORDS:

Charged particles traveling through matter lose energy in the following ways:

In Coulomb interactions with electrons and nuclei. By emission of electromagnetic radiation

(bremsstrahlung). By emission of Cerenkov radiation. Cerenkov

radiation is visible electromagnetic radiation emitted by particles traveling in a medium, with speed greater than the speed of light in that medium. It constitutes a very small fraction of energy loss.