Nuclear Physics Concept Page - 11

Example

Problem on first order disintegration

Example: Nuclei A and B convert into a stable nucleus C. Nucleus A is converted into C by emitting 2 α−$\alpha -$ particles and 3β−$3\beta -$ particles.
Nucleus B is converted into C by emitting one α−particle$\alpha -particle$ and 5 β−particles$\beta -particles$. At time t=0$t=0$, nuclei of A are 4N0$4N_0$ and nuclei of B are N0$N_0$. Initially, number of nuclei of C are zero. Half-life of A (into conversion of C) is 1 min and that of B is 2 min. Find the time (in minutes) at which rate of disintegration of A and B are equal.

Solution:
A→C$A\to C$
B→C$B\to C$
First order decay
dAdt=−λAA$\frac{dA}{dt}=-{\lambda }_{A}A$
dBdt=−λBB$\frac{dB}{dt}=-{\lambda }_{B}B$
A=4Noe−λAt$A=4N_o{e}^{-{\lambda }_{A}t}$
B=Noe−λBt$B=N_o{e}^{-{\lambda }_{B}t}$
rates of disintegration equal
dAdt=dBdt$\frac{dA}{dt}=\frac{dB}{dt}$
4λAe−λAt=λBe−λBt$4{\lambda }_A{e}^{-{\lambda }_{A}t}={\lambda }_B{e}^{-{\lambda }_{B}t}$
4ln21e−ln21t=ln22e−ln22t$4\frac{ln2}{1}{e}^{-\frac{ln2}{1}t}=\frac{ln2}{2}{e}^{-\frac{ln2}{2}t}$
t=6$t=6$ min

Definition

Sources of radiation

Two main sources of radiations are:
(a) Radioactive fall out from nuclear plants and other sources
(b) Disposal of nuclear waste

Definition

Describe Artificial transmutation

Artificial transmutation is the conversion of one element into another by artificial means. It means, changing one element completely by bombarding it with some fundamental particles such as alpha particles, neutrons, deuterons etc. Artificial transmutation is formation of new element by artificial means.

Definition

Change in nucleus due to alpha decay

When a nucleus undergoes alpha-decay, it transforms to a different nucleus by emitting an alpha-particle (a helium nucleus,42He${}_2^{4}He$).
For example, when 23892U${}_{92}^{238}U$undergoes alpha-decay, it transforms to 23490T${}_{90}^{234}T$
23892U→23490Th+42He${}_{92}^{238}U{\to }_{90}^{234}Th{+}_2^{4}He$
In this process, it is observed that since 42He${}_2^{4}He$ contains two protons and two neutrons, the mass number and the atomic number of the daughter nucleus decreases by four and two, respectively. Thus, the transformation of a nucleus AZX${}_Z^{A}X$ into a nucleus A−4Z−2Y${}_{Z-2}^{A-4}Y$ can be expressed as
AZX→A−4Z−2Y+42He${}_Z^{A}X{\to }_{Z-2}^{A-4}Y{+}_2^{4}He$
where AZX${}_Z^{A}X$ is the parent nucleus and A−4Z−2Y${}_{Z-2}^{A-4}Y$ is the daughter nucleus.

Definition

Disintegration energy of a nuclear reaction

The difference between the initial mass energy and the final mass energy of the decay products is called the Q value of the process or the disintegration energy. Thus, the Q value of an alpha decay can be expressed as
Q=(mx−my−mHe)c2$Q=(m_x-m_y-m_{He})c^2$
This energy is shared by the daughter nucleus and the alpha particle,  in the form of kinetic energy.

Definition

Change in nucleus due to beta decay

A nucleus that decays spontaneously by emitting an electron or a positron is said to undergo beta decay.In beta minus (β−${\beta }^{-}$ ) decay, an electron is emitted by the nucleus and a neutron transforms into a proton within the nucleus according to
n→p+e−+ν¯$n\to p+e^{-}+\overline{\nu }$
In beta plus (β+${\beta }^{+}$ ) decay, a positron is emitted by the nucleus and a proton transforms into neutron (inside the nucleus) viap→n+e++ν$p\to n+e^{+}+\nu$

Definition

Examples of beta decay

There are two kinds of beta decay: β−${\beta }^{-}$ and β+${\beta }^{+}$
Examples:

1. β−${\beta }^{-}$ decay:
   3215P→3216S+e−+ν¯${}_{15}^{32}P{\to }_{16}^{32}S+e^{-}+\overline{\nu }$
2. β+${\beta }^{+}$ decay:
   2211Na→2210Ne+e++ν${}_{11}^{22}Na{\to }_{10}^{22}Ne+e^{+}+\nu$

Definition

Ground state

The ground state of a quantum mechanical system is its lowest energy state, the energy of the ground state is known as the zero point energy of the system.

Example

Problem based on gamma decays

In a γ−$\gamma -$decay process, γ−$\gamma -$rays of energy E is emitted. Find the decrease in internal energy of mass M (of nucleus).By momentum conservation
Ec=Mv$\frac{E}{c}=Mv$
v=EMc$v=\frac{E}{Mc}$
Now total decrease in internal energy = Energy of γ$\gamma$ + KE of M
=E+12Mv2$=E+\frac{1}{2}M{v}^2$
=E+E22Mc2$=E+\frac{E^2}{2M{c}^2}$

Definition

Changes within the nucleus in alpha, beta and gamma emission

Alpha emission: If the nucleus of a radioactive element X of mass number A and atomic number Z emits an α$\alpha$ particle, a new element Y (daughter nucleus) is formed which has mass number equal to (A-4) and atomic number equal to (Z-2). Thus due to emission of an alpha particle, atomic number Z decreases by two units and mass number decreases by 4 units.
Beta emission: In emitting a beta particle the number of nucleons in the nucleus (i.e. protons and neutrons) remain same, but the number of nuetrons is decreased by one and the number of protons is increased by one.
Gamma emission: In emitting gamma particle there is no change in mass number A and atomic number Z of the nucleus.

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