Contents
The study of Physics Topics involves the exploration of matter, energy, and the forces that govern the universe.
What is Artificial Radioactivity? Is Radioactive Isotope and Radioisotope the Same?
Definition: When an unstable isotope of a stable element is artificially formed and if this isotope exhibits natural radioac-tivity it is called artificial or induced radioactivity.
Examples: Carbon, sodium, phosphorus are stable elements and their isotopes C12, Na23 and P31 are stable isotopes. When C14, Na24, P30 isotopes are produced artificially, they are found to be radioactive. They are generally called radioactive isotopes or radioisotopes. Like C14 is called radiocarbon, Na24 is known as radiosodium and so on.
Radioactive decay mode of these isotopes can be represented as
6C14 → 7N14 + -1β0 ; half-life = 5600 y
11Na24 → 12Mg24 + -1β0; half-life = 15 h
15P30 → 14Si30 + +1β0; half-life = 2.5 min
Positive β decay or β+ decay: As evident from above radioactive decay of P30, the emitted particle is +1β0 or positron. Positrons have same mass as electrons but are positively charged (+e). Except for the positive charge, the β-decay and positron emission are identical. So, it is called β+ decay. β+ decay is found only in artificial radioactivity.
Characteristics of artificial radioactivity:
- Radioisotopes are produced from elements which are natu-rally stable.
- Decay mode of radioisotopes is similar to natural radioac-tivity of radioactive substances. It follows exponential law of radioactivity.
- Natural radioactive substances generally have higher mass number. But radioisotopes may be lighter.
- Radioisotopes can exhibit α -decay, β -decay, γ -decay or β+ decay, last one being characteristic of artificial radioac-tivity only.
- The displacement rule for β+ decay is that mass number remains unaltered but atomic number decreases by 1.
Discovery of Curie-Joliot: Irine Curie and Frederic Joliot first discovered artificial radioactivity. Projecting a -particles from polonium on aluminium target, they identified positron particles mixed with emitted neutrons. While emission of neutron is easily explained using the nuclear reaction
2He4 + 13Al27 → 15P30 + 0n1 the presence of positrons remained unexplained.
They further observed that
i) On stopping projectile α from the source, emission of neutron is stopped but emission of positrons continue.
ii) The rate of decrease of positron emission from Al -target excited by α -particles, is exponential, i.e., it obeys the decay law of natural radioactivity.
They, therefore, concluded that isotope P30 produced in the above nuclear reaction must be a radioactive isotope.
Neutron Induced Nuclear Reactions
Neutron, as a projectile in bringing about a nuclear reaction, has definite advantages over α -particles or protons as projectiles. Being a neutral particle, initial energy of a neutron is of little sig-nificance as it does not have to overcome the electrostatic repul-sion from the positive nucleus. Whereas α -particles or protons should be of high energy, to overcome the coulomb barrier to reach the nucleus, neutrons are ideal projectiles for a nuclear reaction, as they do not have to lose any energy to reach the nucleus.
Thermal neutron: Nuclear reactions brought about by neutrons of energy of a few MeV are similar to reactions caused by using α or proton projectiles. But when a slow moving neutron hits a target, a set of new types of nuclear reactions take place.
Neutrons of kinetic energy of the order of 10-2 eV are called thermal neutrons. This order is the same as the kinetic energy of atomic or subatomic particles at room temperature. Since, external manifestation of kinetic energy of atomic or subatomic particles is the thermal energy, these are called thermal neutrons. Thermal neutrons are actually very slow moving neutrons.
U-238 and U-235 : Two isotopes of uranium are present in the ore of natural uranium. These are 92U238 and or U-238 and U-235 for short. Their abundance in nature is in the ratio 140 : 1 (99.28% : 0.7%).
Transuranic elements: Atomic number of uranium is 92 and no element of atomic number higher than 92 is found in nature. But when U-238 is hit by a thermal neutron then,
i) the neutron is absorbed by U-238 and U-239 is formed and
ii) U-239 undergoes decay and forms an isotope of atomic number 93 , which is a new element and does not exist naturally. This artificially made element is called neptunium (Np).
- 0n1 + 92U238 → 92U238
- 92U239 → 93Np239 + -1β0
Proceeding almost in the similar way, it has been possible to produce elements of atomic numbers 94, 95, 96, ……., 118 in the laboratory. Elements of atomic number higher than 92 and produced artificially are called transuranic elements. Out of these elements plutonium (94Pu239) has the maximum practical use.
List of a few natural and artificial isotopes, decay mode and half-life period: (Note that associated γ -radiation not shown.)
Numerical Examples
Example 1.
Complete the following nuclear reaction:
13Al27 + 2He4 → 15P30 + ?
Solution:
Let A and Z be mass and atomic number of the unknown particle respectively. From the law of conservation of mass number
27 + 4 = 30 + A or, A = 1
From the law of conservation of atomic number,
13 + 2 = 15 + Z or, Z = 0
The particle is therefore a neutron 0n1.
∴ The complete equation of the reaction:
13Al27 + 2He4 = 15P30 + 0n1
Example 2.
Complete the following nuclear reaction:
7N14 + 2He4 → 8O17 + ?
Solution:
According to the law of conservation of mass number,
14 + 4 = 17 + 1
Again, according to the law of conservation of atomic number,
7 + 2 = 8 + 1
∴ The mass number of the unknown particle = 1 and its atomic number = 1.
So, it is proton.
∴ The complete equation of the reaction:
7N14 + 2He4 → 8O17 + 1H1
Example 3.
Identify the missing particle in the following two reactions
(i) 9F19 + 1H1 → 8O16 + ?
(ii) 12Mg25 + ? → 11Na22 + 2He4
Solution:
i) Since, 19 + 1 = 16 + 4 and 9 + 1 = 8 + 2, the mass number of the missing element = 4 and atomic number = 2. As Z = 2, the element is α -particle.
∴ The complete equation of the reaction:
9F19 + 1H1 → 8O16 + 2He4
ii) As, 25 + 1 = 22 + 4 and 12 + 1 = 11 + 2, the mass number of the missing element = 1 and atomic number = 1.
So, the element is proton.
∴ The complete equation of the reaction:
12Mg25 + 1H1 → 11Na22 + 2He4
Example 4.
When 4Be9 is hit by α -particles a neutron is emitted resulting in the formation of a new element. Identify the element and write the complete reaction equation.
Solution:
α -particle: 2He4; neutron: 0n1
∴ Let the new element be
From the laws of conservation of mass number and atomic number,
9 + 4 = A + 1 or, A = 12
and 4 + 2 = Z + 0 or, Z = 6
∴ The new element = 6C12 (carbon).
∴ The complete equation of the reaction:
4Be9 + 2He4 → 6C12 + 0n1
Example 5.
When an aluminium nucleus (13Al27) is hit by a proton a new element is formed with the emission of α -particle.
(i) Write the complete equation of reaction;
(ii) Identify the new element and
(iii) Determine the number of neutrons and protons in the nucleus.
Solution:
Proton: 1H1; α -particle: 2He4
∴ Let the new element = ZXA
[where A = mass number; Z = atomic number]
From the laws of conservation of mass and atomic number,
27 + 1 = A + 4 and 13 + 1 = Z + 2
or, A = 24 and Z = 12
i) The complete equation of the reaction:
13Al27 + \({ }_1^1 \mathrm{H}\) → 12Mg24 + 2He4
ii) As Z = 12, the element is magnesium (Mg).
So, new element = 12Mg24.
iii) Now, we know, mass number = proton number + neutron number (x)
or, 24 = 12 + x [∵ proton number = atomic number
or, x = 12
Example 6.
On collision with neutron, 13Al27 changes to radio sodium 11Na24 and emits a particle. 11Na24, in its turn emits a particle and is transmuted to 12Mg24. Write the two nuclear equations and identify the particles.
Solution:
Let the first equation of the reaction be
13Al27 + 0n1 → 11Na24 + ZXA ….. (1)
and the second equation of the reaction be
11Na24 → 12Mg24 + \(z_1 X^{A_1}\) …. (2)
Applying the laws of conservation of atomic number and mass number,
from equation (1), 27+ 1 = 24 + A or, A = 4
and 13 + 0 = 11 + Z or, Z = 2
Also, from equation (2), 24 = 24 + A1 or, A1 = 0
and 11 = 12 + Z1 or, Z1 = -1
Hence, the particles are helium nucleus i.e., α -particle in equation (1) and β-particle in equation (2).
The complete equations are
13Al27 + 0n1 → 11Na24 + 2He4
and 11Na24 → 12Mg24 + -1β0
Example 7.
A nucleus disintegrates into two nuclei and their velocities are in the ratio of 2: 1 . What will be the ratio of their sizes? (AIEEE 04)
Solution:
From the law of conservation of momentum, the two nuclei will have the same magnitude of momentum. Hence, mass ∝ \(\frac{1}{\text { velocity }}\). Again, mass ∝ R3, where R is the radius.
∴ mass ∝ R3 ∝ \(\frac{1}{\text { velocity }}\) ∴ \(\frac{R_1}{R_2}\) = \(\left(\frac{v_2}{v_1}\right)^{\frac{1}{3}}\) = \(\left(\frac{1}{2}\right)^{\frac{1}{3}}\)
∴ R1 : R2 = 1 : 21/3
Example 8.
In the nuclear reaction X(n, α)3Li7, Identify X. [AIEEE ‘05]
Solution:
The equation of the reaction:
ZXA + 0n1 → 3Li7 + 2He4
Using the conservation laws,
A + 1 = 7 + 4 or, A = 10
and Z + 0 = 3 + 2 or, Z = 5
These are the Z and A values of boron (B).
∴ The unknown element, X = 5B10 (boron nuclide)