Table of contents | |
Alpha Decay | |
Beta Decay | |
Gamma Decay | |
Activity |
Nuclei which contain 210 or more nucleons are so large that the short range nuclear forces that hold them together are barely able to counterbalance the mutual repulsion of their protons. Alpha decay occurs in such nuclei as a means of increasing their stability by reducing their size
To escape from nucleus, a particle must have K.E., and only the alpha particle mass is sufficiently smaller than that of its constituent nucleons for such energy to be available (α-particle have high B.E. as compared to proton or nuclei).
The energy Q-released when various particles are emitted by a heavy nucleus is, i.e.
Disintegration energy Q = (mi - mf - mx)c2 where
mi = Mass of initial nuclei,
mf = mass of final nuclei,
mx = α -particle mass
The KEα of the emitted α-particle is never quite equal to Q, since momentum must be conserved, the nucleus recoils with a small amount of kinetic energy when the α-particle emerges. Thus
since A ≥ 210, most of the disintegration energy appears as the K.E. of the α-particle.
Although α-decay is inexplicable classically, quantum mechanics provides a straight forward explanation. The basic notions of this theory are:
An α-particle may exist as an entity within a heavy nucleus. Such a particle is in constant motion and is held in the nucleus by potential barrier. There is a small but definite-likelihood that the particle may tunnel through the barrier (despite its height) each time a collision with it occurs.
The decay probability per unit time, i.e decay constant λ = vT
Where v = number of times per second an -particle within a nucleus strikes the potential barrier around it and = Probability that the particle will be transmitted through the barrier.
L is the width of the barrier, wave number inside the barrier where E is the K.E. , U is height of the barrier and m is the mass of α-particle.
It is a means whereby a nucleus can alter its composition to become more stable. The conservation principles of energy, linear momentum and angular momentum are all apparently violated in beta decay:
n → p + e-
(i) The electron energies observed in the β- decay of a particular nuclide are found to vary continuously from 0 to maximum value max KEmax characteristic of the nuclide. The maximum energy Emax = (m0c2 + KEmax) carried by the decay electron is equal to the energy equivalent of the mass difference between the parent and daughter nuclei. However an emitted electron is rarely found with energy of KEmax.
(ii) When the directions of emitted electron and of the recoiling nuclei are observed, they are almost never exactly opposite as required for linear momentum to be conserved.
(iii) The spins of the neutron, proton and electron are all 1/2. If beta decay involved just a neutron becoming a proton and an electron, spin (and hence angular momentum) is not conserved.
In 1930 Pauli proposed a "desperate remedy": If an uncharged particle of small or zero rest mass and spin 1/2 is emitted in β- - decay together with the electron, the above discrepancies would not occur. This particle is called neutrino which would carry off energy equal to the difference between KEmax and actual K.E of the electron (the recoiling nucleus carry away negligible K.E). The neutrino's linear momentum also exactly balances those of the electron and the recoiling daughter nucleus.
Thus in ordinary β- - decay (also possible outside the nucleus)
The interaction of neutrinos with matter is extremely feeble. The only interaction with matter a neutrino can experience is through a process called inverse beta decay with extremely low probability
Note: Parity violates in β- - decay.
γ-rays characteristics
1. It is an electromagnetic wave.
2. Very short wavelength (≈400Ao to 0.4Ao).
3. No electric charge and so not detected by magnetic and electric field.
When a beam of γ-rays photons passes through matter, the intensity of beam decreases exponentially i.e. I = I0e-μx where I0 : Initial Intensity, µ: absorption coefficient of substance, x : thickness of absorber.
Pair Annihilation
The inverse of pair production occurs when a positron is near an electron and the two come together under the influence of their opposite electric charges. Both particles vanish simultaneously with the lost mass becoming energy in the form of two gamma ray photon.
e+ + e- → γ + γ
The total mass of the positron and electron is equivalent to 1.02 MeV , and each photon has energy h of 0.51 MeV plus half the K.E. of the particles relative to their center of mass.
Note:
1. The directions of the photons are such as to conserve both energy and linear momentum.
2. No nucleus or other particles is needed for this pair annihilation to take place.
A given type of of nuclei will normally decay by one particular mode; say by emission of β-particles. But many cases have been found in which a smaller percentage of nuclei will decay by a different mode such as α-emission.
Let us denote the probability of α-emission by one nucleus, in dime dt emission by λαdt and that of β-emission by λβdt.
Then the probability of decay of a nucleus in time dt by either α or β-emission is: (λα + λβ)dt.
Hence the activity is dN/dt
Giving mean life and Branching Ratio λα/λβ
(Stable Product, λpb = 0)
The half life of U238 is 4.5 x 109 Years. Hence after sufficient time the only element present in any appreciable amount will be uranium and lead.
Here λ1 = λU , λ2 = λPb = 0 , N2 = NPb and N0 = Nv
Thus
N0 = Nv = Present no. of Pb atoms+Present no. of U atoms ⇒ Nv = NPb + NU
Example: Half life of P is 14.3 days. If you have 1.00 g of P today, then what would be the amount remaining in 10 days
N = Noe-λt ⇒ N = (1.00)e-0.04847x10 ⇒ N = 0.616 g or N = 616mg
Example: A radioactive nucleus has a half life of 100 years. If the number of nuclei t = 0 is N0 , then find the number of nuclei that have decayed in 300 years.
Number of nuclei present after 300 year
Example: The atomic ratio between the uranium isotopes 238U and 234U in a mineral sample is found to be 1.8 x 104. The half life of 238U is 4.5 x 109 years , then find the half life of 234U.
Example: A radioactive sample contains 1.00 mg of radon 222Rn , whose atomic mass is 222 u. The half life of the radon is 3.8 day. Then find the activity of the radon.
Number of atoms in 1.00 mg is
Hence, activity R = λN = 2.1x10-6 x 2.7 x1018 = 5.7 x 1012 decay / sec
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1. What is alpha decay? |
2. What is beta decay? |
3. What is gamma decay? |
4. What is activity in the context of radioactive decay? |
5. How is radioactive decay relevant to the IIT JAM exam? |
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