Contents
Physics Topics cover a broad range of concepts that are essential to understanding the natural world.
Define 1 curie? What is Nuclear Reaction?
Definition: The rate of radioactive disintegration with time is called the activity of the sample.
Mathematically, activity (A) = \(\frac{d N}{d t}\) = λN, the numerical value.
Therefore, activity A ∝ N and A ∝ λ ∝ \(\frac{1}{T}\).
From this, we can say, a radioactive sample has greater activity if
- the sample contains large number of radioactive atoms (N)
- decay constant is high or half-life is low.
Also if A0 and A are the activities initially and after a time t, then
A0 = λN0, and A = λN
∴ \(\frac{A}{A_0}\) = \(\frac{N}{N_0}\) = e-λt [using N = N0e-λt]
or, A = A0e-λt
∴ Activity also decreases exponentially with time.
Units to measure activity: The activity of a radioactive substance is measured in terms of number of disintegration per unit time. In SI, the unit is becquerel or Bq and 1 Bq = 1 dis-integration per second or 1 dps.
Other practical units are
curie : 1 Ci = 3.70 × 1010 dps
rutherford : 1 Rd = 106 dps
Numerical Examples
Example 1.
Po210 has half-life of 140 d. In 1 g Po210 how many disintegration will take place every second? [Avoga- dro’s number = 6.023 × 1023]
Solution:
Disintegration constant,
λ = \(\frac{0.693}{T}\) = \(\frac{0.693}{140 \times 24 \times 60 \times 60}\)s-1
Number of atoms in 210 g Po210
= Avogadro’s number = 6.023 × 1023
∴ Disintegration per second = activity = λN
= \(\frac{0.693}{140 \times 24 \times 60 \times 60}\) × \(\frac{6.023 \times 10^{23}}{210}\)
= 1.64 × 1014 dps
Example 2.
A radioactive sample of half-life 30 d contains 1012 particles at an instant of time. Find the activity of the sample.
Solution:
Decay constant,
λ = \(\frac{0.693}{T}\) = \(\frac{0.693}{30 \times 24 \times 60 \times 60}\)s-1 and N = 1012
∴ Activity = A = λN = \(\frac{0.693 \times 10^{12}}{30 \times 24 \times 60 \times 60}\) = 2.67 × 105 dps
Example 3.
How much 84Po210 of half-lIfe 138 days is required to produce a source of α -radiation of intensity 5 mCi (millicurie)?
Solution:
Decay constant, λ = \(\frac{A}{\lambda}\) = \(\frac{5 \times 3.7 \times 10^7 \times 138 \times 24 \times 60 \times 60}{0.693}\)s-1
Activity, A = 5 mCi = 5 × 3.70 × 107 dps
Now, A = λN
or, N = \(\frac{A}{\lambda}\) = \(\frac{5 \times 3.7 \times 10^7 \times 138 \times 24 \times 60 \times 60}{0.693}\)
Again, number of atoms contained in 210 g P210
= Avogadro’s number = 6.023 × 1023
Hence, mass of N such particles
= \(\frac{210 \times 5 \times 3.7 \times 10^7}{6.023 \times 10^{23}}\) × \(\frac{138 \times 24 \times 60 \times 60}{0.693}\)
= 1.11 × 10-6 g (approx.)
Example 4.
A 280 days old radioactive substance shows an activity of 6000 dps, 140 days later It’s activity becomes 3000 dps. What was its initial activity? [IIT ‘04]
Solution:
In the table, the last two values of activity are given. These are used to calculate the first two values.
Hence, initial activity = 24000 dps
Artificial Transmutation of Elements
We have already seen that radioactive elements are transformed into new elements. We also know that the identity of an element depends on its proton number. Therefore, if the proton number of an element can somehow be changed, the element is said to have undergone artificial transmutation.
Generally, artificial transmutation is brought about in two ways—
1. Nuclear reaction: Here the nucleus is bombarded with high energy particles. Thereby the nucleus undergoes a change and forms the nucleus of a new element [discussed in Section 2.12.1].
2. Artificial radioactivity: Often the transmuted nucleus formed by the process of nuclear reaction, is not a stable one and exhibits radioactivity and thereby decays to form a stable nucleus of another element, i.e., a transmuted element. Note that only the nuclear reaction is artificial and the subsequent disintegration of the unstable radioactive product is a natural phenomenon [discussed in Section 2.12.2].
Nuclear Reactions
Definition: Nuclear reaction is the process of transmuting elements by bringing about a change in the nucleus, artificially.
Energy condition of nuclear reactions: Binding energy per nucleon of a stable nucleus is about 8 MeV. Almost equal or more than this amount of energy is to be supplied from outside to bring about any change in the nucleus. Generally, high energy particles are used to hit the nucleus to supply energy and to break up the nucleus. The sources of these high energy particles are –
- α, β and γ -rays from radioactivity,
- stream of high energy particles resulting from nuclear reactions can be used to bring about further nuclear reaction,
- particles as projectiles from particle accelerators like cyclotron, betatrons etc. particles as projectiles from particle accelerators like cyclotron, betatrons etc.
Equations of nuclear reactions: In any nuclear reaction, a still or stationary substance called target is hit by a stream of high energy particles, each called a projectile. When a projectile hits the target then either of the two things happens:
i) Target and projectile both remain unaltered. This process is called scattering.
ii) The nucleus of the target changes into the nucleus of another element. The impact produces one or more new particles emerging with high energy. These particles are called emergent particles and the newly formed nucleus is called product nucleus. This entire process is called a nuclear reaction.
Hence, denoting the nucleus of the target as X, projectile as a, product nucleus as Y and emergent particle as b, the nuclear reaction can be shown as
a + X → y + b …. (1)
Conservation of mass number: If A1, A2, A3 and A4 are mass numbers of a, X, Y and b, then from equation (1)
A1 + A2 = A3 + A4 ……… (2)
Conservation of atomic number: If Z1, Z2, Z3 and Z4 are atomic numbers of a, X, Y and b, then from equation (1)
Z1 + Z2 = Z3 + Z4 …… (3)
Mass-energy conservation: Mass lost during the nuclear reaction, changes to energy as per Einstein’s mass-energy equivalence. The released energy in a nuclear reaction is called Q -value of the reaction. Q -value for particles from the equation (1),
Q = [(Ma + MX) – (MY + Mb)]c2 ….. (4)
The reaction is exoergic when Q -value is positive and endoergic when Q -value is negative. In the second case the reaction cannot take place unless a threshold energy is provided to the projectile.
Rutherford’s experiment : Proton as a nuclear particle: Rutherford was the first to bring about artificial transmutation of elements. His experimental arrangement is represented schemetically, in Fig.
C is a container with a window W covered by a thin sheet of aluminium. F is a detector or film to record any emergent particle through W. R is a source of polonium that emits α -rays in the decay process. It is placed opposite to the window. Chamber C is filled with pure nitrogen gas. The observations after a considerable period of time are:
- Chamber C shows the presence of oxygen gas, on chemical analysis.
- Photographic plate F is exposed and relevant detectors establish that the emerging rays through the window are streams of high energy protons.
From this experiment it is understood that proton is one of the constituents of atomic nucleus.
Detailed investigations showed that proton and hydrogen nucleus were identical. Hence, symbol for proton in nuclear reaction is 1H1.
Equation of reaction: The nuclear reaction in the chamber C, can be represented by
In short form the reaction is often represented as N14(α, p) O17 and the explanation is—when N14 is bom-barded with an α -particle (2He4), O17 (oxygen) is produced and a proton is emitted.
A few α-induced transformations:
- 2He4 + 13Al27 → 14Si30 + 1H1 or, Al27(α, p) Si30
- 2He4 + 5B10 → 6C13 + 1H1 or, B10(α, p) C13
- 2He4 + 19K39 → 20Ca42 + 1H1 or, K39(α, p) Ca42
Discovery of neutron : Chadwick’s experiment
Bothe-Baker’s experiment: These German scientists in 1930, observed that when Be (beryllium) is exposed to a stream of α -particles, highly penetrating uncharged rays, are emitted. Initially, these rays were considered γ -rays.
Curie-Joliot’s experiment: In 1932, Irine Curie and her husband Frédéric Joliot observed that when a block of paraffin wax was placed on the path of above mentioned rays, high energy protons were emitted. Emissions in Bothe and Becker experiment, if taken as γ -rays, cannot account for the source of the high energy protons produced in this experiment.
Chadwick’s Analysis: James Chadwick in the same year repeated the experiment and put forward the explanation. His experimental arrangement is shown in Fig.
Chadwick assumed that
i) The rays emitted due to the impact of α -particles on Be nucleus, were not electromagnetic waves like γ -rays but a stream of neutral particles. He named these particles neutron.
ii) Paraffin wax contains hydrogen atoms and every nucleus of the atom is a proton. Due to the elastic collision between the stream of neutrons and the protons, the proton stream resulted.
Ionisation chamber helps in finding the energy and momentum of the proton released. Now, on applying the theory of elastic collision the mass of neutron can be obtained.
This experiment established neutron as a fundamental particle that constitutes a nucleus. Three things are inferred here.
i) That neutron is electrically neutral. Its mass is slightly greater than the mass of proton. So its effective atomic number and mass number are 0 and 1 respectively. So in a nuclear reaction it is represented as 0n1.
ii) Free neutron is not a stable particle. It undergoes a natural β -decay and changes to a proton.
0n1 → 1H1 + -1β0 (electron)
The half-life period for this radioactivity is about 12 min.
iii) The nuclear reaction in beryllium can be represented as
2He4 + 4Be9 → 6C12 + 0n1