NUCLEAR AND PARTICLE PHYSICS -4

                                            Nuclear Decay

s

The nuclear decay can be divided into three categories

α-decay: spontaneous emission of α-particle from a nucleus of large atomic number. This sets an upper limit of atomic numbers of chemical elements occurring in nature. 

β-decay: spontaneous emission or absorption of an electron or positron by a nucleus. This tells about the nature of forces.

γ-decay: spontaneous emission of high energy photons when nucleus makes transition from excited state to ground state..

Velocity and Energy of alpha-particles (Doubly charged ₂He⁴):

1.  Determination of velocity and energy of α-particles made it possible to determine energies which differ only in small amounts and this led to the discovery that some radionuclides actually emit spectrum of α-particles.

2.  Knowledge of energies helped to assign certain nuclear energy levels with confidence.

3.  The method of determining the energies of α-particles are also used for protons and neutrons. When a charged particle moves in a magnetic field, its orbit is a circle of radius r given by 

𝐻𝑞𝑣 = Mv²/r

H- field strength, q and M are the charge and mass of the particle.

𝐻𝑞𝑟

𝑣 = _______ 

𝑀

v is determined by knowing H and r.                    q/M is known.

q = 3.2043×10^-20 emu, M = 6.6430×10^-24 gm. q/M = 4823.5 emu/gm

v = 4823.5 H r,   H~10,000 Gauss

H r = 3,00,000 to 5,00,000 Gauss cm.

𝐸 = (1/2) 𝑀𝑣² =(1/2) × 6.6430 × 10⁻²⁴ × 𝑣² 𝑒𝑟𝑔𝑠 = 3.3215 × 10⁻²⁴𝑣²𝑒𝑟𝑔𝑠

Relativistic correction gives

                                                                                             

                                                                                      

                                                           

Range, Ionization and Stopping Power:

Alpha particles are absorbed by a sheet of paper, aluminium foil of 0.004 cm thick.

When they travel through matter they lose energy due to collisions with particles of matter (air or gas). They produce ion-pairs in air. α-particles are absorbed after travelling a certain distance in the medium. This distance up to which α-particles travel into the medium prior to being absorbed by the medium is known as the range of the α-particles. 

 

 

The straight line portion AB is extrapolated. Extrapolated value R_e when the intensity is half the initial intensity, the distance of the particle from the source is mean range R.

https://youtu.be/53TIyqIMdYA

Measurement of Range of α-particles:

S-source of α-particles placed on movable platform.   P is plate and G is the grid . A potential difference is applied and connected to electrometer. The ionization produced by α-particles at different distance from S was determined by the ionization chamber. The specific ionization against distance from source is shown which is called Bragg’s curve.                                                                

Specific ionization remains almost constant, then attains maximum and falls to zero. Tail is due to the phenomenon of straggling. Some α-particles may absorb one electron and He⁺ still continues to travel producing ionization and may absorb one more electron to become neutral.

Stopping Power:  Stopping power is defined as the energy lost by α-particle per unit distance as it travels through the medium.

                                                  S(E) = ^{𝑑𝐸𝛼 /dx}

𝑅𝑎𝑛𝑔𝑒 𝑜𝑓 α−particles  in air

Relative stopping power (RSP) = ________________________________________     

𝑅𝑎𝑛𝑔𝑒 𝑜𝑓 α−particles in material

 

α –disintegration energy: 

When α-particle is emitted, the product nucleus recoils carrying certain amount of energy. Both momentum and energy are conserved

.^𝐴_𝑍𝑋 → ^𝐴_𝑍⁻₋⁴₂𝑌 + ⁴₂𝐻𝑒 (α − particle)

²³⁸92𝑈 → ²³⁴₉₀𝑇ℎ + ⁴₂𝐻𝑒 + 𝐸_𝛼(𝑑𝑖𝑠𝑖𝑛𝑡𝑒𝑔𝑟𝑎𝑡𝑖𝑜𝑛 𝑒𝑛𝑒𝑟𝑔𝑦)

Where E_α is the sum of the K.E. of α-particle and product nucleus.

From conservation of momentum,

Mv = M_αv_{α ;}    M_α,  v_α are the mass and velocity of α-particle; M, v are mass and velocity of product nucleus.

𝐸𝛼 = 21 𝑀𝛼𝑣𝛼2 + 12 𝑀𝑣2 

𝐸𝛼 = 12 𝑀𝛼𝑣𝛼₂(1 + 𝑀𝑀𝛼𝑣𝑣²𝛼₂ )

1

𝐸𝛼 = 2 𝑀𝛼𝑣𝛼2(1 + 𝑀𝑀𝛼22𝑣𝑣𝛼22𝑀𝑀𝛼) 

𝐸𝛼 = 21 𝑀𝛼𝑣𝛼2(1 + 𝑀𝑀𝛼) 

𝐸_𝛼 = 𝐾_𝛼(1 + 𝑀_𝑀^𝛼) where K_α =  ¹₂ 𝑀_𝛼𝑣_𝛼²   is the K.E. of α-particle

𝑀

𝑇ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒, 𝐾_𝛼 = 𝐸_𝛼(_𝑀+𝑀𝛼) 

Kinetic energy of product nucleus is

K= Eα - Kα

                                                                          𝑀                                               𝑀 + 𝑀_𝛼 − 𝑀

𝐾 = 𝐸_𝛼 − ___________                  ;𝐸_𝛼 = 𝐸_𝛼(___________________-)

                                                                    𝑀 + 𝑀_𝛼                                                            𝑀 + 𝑀_𝛼

𝑀^𝛼 

     ______                                        Mass of parent  nucleus = A amu

          𝐾 = 𝐸_𝛼(_𝑀+𝑀𝛼)

𝐾_𝛼 = 𝐸_𝛼(M/(_𝑀+𝑀𝛼))                                           Mass of product nucleus = A - 4 amu

= 𝐸_𝛼(A-4/ A)                                                    M_α + M = A

 E_α = (M_i – M_f – M_α) × c²                            M = A - M_α = A - 4

  = 931 ×  (M_i – M_f – M_α) MeV            M_i = initial mass (amu), M_f = final mass (amu)

²³⁸92𝑈 → ²³⁴₉₀𝑇ℎ + ⁴₂𝐻𝑒 + 𝐸_𝛼            

E_α5.40 MeV 

E_α is the total energy released in the decay process and is called disintegration energy.

 

biswatutorialphysics.blogspot.com

biswatutorialphysics.blogspot.com