Topic Covered

`color{blue}{star}` RADIOACTIVITY


`color{brown}bbul("Mass – Energy : ")`
`color{blue} ✍️`Einstein showed from his theory of special relativity that it is necessary to treat mass as another form of energy. Before the advent of this theory of special relativity it was presumed that mass and energy were conserved separately in a reaction.

`color{blue} ✍️`However, Einstein showed that mass is another form of energy and one can convert mass-energy into other forms of energy, say kinetic energy and vice-versa. Einstein gave the famous mass-energy equivalence relation

`color{navy}(E = mc^2)`

.......... (13.6)

`color{blue} ✍️`Here the energy equivalent of mass m is related by the above equation and c is the velocity of light in vacuum and is approximately equal to `color{purple}(3xx10^8ms^(-1)).`

`color{blue} ✍️`Experimental verification of the Einstein’s mass-energy relation has been achieved in the study of nuclear reactions amongst nucleons, nuclei, electrons and other more recently discovered particles.

`color{blue} ✍️`In a reaction the conservation law of energy states that the initial energy and the final energy are equal provided the energy associated with mass is also included. This concept is important in understanding nuclear masses and the interaction of nuclei with one another. They form the subject matter of the next few sections.

`color{brown}bbul("Nuclear binding energy")`
`color{blue} ✍️`In previous theory, we have seen that the nucleus is made up of neutrons and protons. Therefore it may be expected that the mass of the nucleus is equal to the total mass of its individual protons and neutrons.

`color{blue} ✍️`However, the nuclear mass `M` is found to be always less than this. For example, let us consider `text()_(8)^(16)O` nucleus which has 8 neutrons and 8 protons. We have

`color{green}("Mass of 8 neutrons = 8 × 1.00866 u")`
`color{green}("Mass of 8 protons = 8 × 1.00727 u")`
`color{green}("Mass of 8 electrons = 8 × 0.00055 u")`

`color{blue} ✍️`Therefore the expected mass of `color{purple}(text()_(8)^(16)O "nucleus" = 8 × 2.01593 u = 16.12744 u.)`

`color{blue} ✍️`The atomic mass of `text()_(8)^(16)O` found from mass spectroscopy experiments is seen to be 15.99493 u. Substracting the mass of 8 electrons (8 × 0.00055 u) from this, we get the experimental mass of `text()_(8)^(16)O` nucleus to be 15.99053 u.

`color{blue} ✍️`Thus, we find that the mass of the `text()_(8)^(16)O` nucleus is less than the total mass of its constituents by 0.13691u.

`color{blue} ✍️`The difference in mass of a nucleus and its constituents, `DeltaM,` is called the mass defect, and is given by

`color{navy}(DeltaM = [Zm_P+(A-Z)m_n]-M)`


`color{blue} ✍️`It is here that Einstein’s equivalence of mass and energy plays a role. Since the mass of the oxygen nucleus is less that the sum of the masses of its constituents (8 protons and 8 neutrons, in the unbound state), the equivalent energy of the oxygen nucleus is less than that of the sum of the equivalent energies of its constituents. If one wants to break the oxygen nucleus into 8 protons and 8 neutrons, this extra energy `DeltaMc^2` has to supplied.

`color{blue} ✍️`This energy required `E_b` is related to the mass defect by

`color{navy}(E_b= DeltaMc^2)`


`color{blue} ✍️`If a certain number of neutrons and protons are brought together to form a nucleus of a certain charge and mass, an energy `E_b` will be released in the process. The energy `E_b` is called the binding energy of the nucleus. If we separate a nucleus into its nucleons, we would have to supply a total energy equal to `E_b`, to those particles.

`color{blue} ✍️`Although we cannot tear apart a nucleus in this way, the nuclear binding energy is still a convenient measure of how well a nucleus is held together.

`color{blue} ✍️`A more useful measure of the binding between the constituents of the nucleus is the binding energy per nucleon, `E_(bn)`, which is the ratio of the binding energy `E_b` of a nucleus to the number of the nucleons, A, in that nucleus:

`color{navy}(E_(bn) = E_b / A)`

........ (13.9)

`color{blue} ✍️`We can think of binding energy per nucleon as the average energy per nucleon needed to separate a nucleus into its individual nucleons.

`color{blue} ✍️`Figure 13.1 is a plot of the binding energy per nucleon `E_(bn)` versus the mass number A for a large number of nuclei. We notice the following main features of the plot:

`color{blue} {(i)}` the binding energy per nucleon, `E_(bn),` is practically constant, i.e. practically independent of the atomic number for nuclei of middle mass number ( 30 < A < 170). The curve has a maximum of about 8.75 MeV for A = 56 and has a value of 7.6 MeV for A = 238.

`color{blue} {(ii)}` `E_(bn)` is lower for both light nuclei (A<30) and heavy nuclei `(A>170).`

`color{blue} ✍️`We can draw some conclusions from these two observations:

`color{blue} {(i)}` The force is attractive and sufficiently strong to produce a binding energy of a few MeV per nucleon.

`color{blue} {(ii)}` The constancy of the binding energy in the range `30 < A < 170` is a consequence of the fact that the nuclear force is short-ranged.

`color{blue} ✍️`Consider a particular nucleon inside a sufficiently large nucleus. It will be under the influence of only some of its neighbours, which come within the range of the nuclear force. If any other nucleon is at a distance more than the range of the nuclear force from the particular nucleon it will have no influence on the binding energy of the nucleon under consideration.

`color{blue} ✍️`If a nucleon can have a maximum of p neighbours within the range of nuclear force, its binding energy would be proportional to p. Let the binding energy of the nucleus be pk, where k is a constant having the dimensions of energy.

`color{blue} ✍️` If we increase A by adding nucleons they will not change the binding energy of a nucleon inside. Since most of the nucleons in a large nucleus reside inside it and not on the surface, the change in binding energy per nucleon would be small. The binding energy per nucleon is a constant and is approximately equal to pk.

`color{blue} ✍️`The property that a given nucleon influences only nucleons close to it is also referred to as saturation property of the nuclear force.

`color{blue} {(iii)}` A very heavy nucleus, say A = 240, has lower binding energy per nucleon compared to that of a nucleus with A = 120.

`color{blue} ✍️`Thus if a nucleus A = 240 breaks into two A = 120 nuclei, nucleons get more tightly bound. This implies energy would be released in the process. It has very important implications for energy production through fission, to be discussed later in Section 13.7.1.

`color{blue} {(iv) }` Consider two very light nuclei `(A =<10)` joining to form a heavier nucleus. The binding energy per nucleon of the fused heavier nuclei is more than the binding energy per nucleon of the lighter nuclei.

`color{blue} ✍️`This means that the final system is more tightly bound than the initial system. Again energy would be released in such a process of fusion. This is the energy source of sun, to be discussed later in Section 13.7.3.
Q 3202723638

Calculate the energy equivalent of 1 g of substance.
Class Chapter 13 Example 2

`E = 10^(–3) × ( 3 × 10^8)^2 J`

`E = 10^(–3) × 9 × 10^16 = 9 × 10^13 J`
Thus, if one gram of matter is converted to energy, there is a release of enormous amount of energy.
Q 3109278118

Find the energy equivalent of one atomic mass unit, first in Joules and then in MeV. Using this, express the mass defect of `text()_(8)^(16)O` in `MeV//c^2.`
Class 12 Chapter 13 Example 3

`1u = 1.6605 × 10^(–27) kg`
To convert it into energy units, we multiply it by `c^2` and find that energy equivalent `= 1.6605 × 10^(–27) × (2.9979 × 10^8)^2 kg m^2//s^2`
`= 1.4924 × 10^(–10) J`
`= (1.4924 × 10^(–10))/(1.602xx10^(-19))eV`
`= 0.9315 × 10^9 eV`
`= 931.5 MeV`
or, `1u = 931.5 MeV//c^2`
For `text()_(8)^(16)O DeltaM= = 0.13691 u = 0.13691×931.5 MeV//c^2`
`= 127.5 MeV//c^2`

The energy needed to separate `text()_(8)^(16)O` into its constituents is thus `127.5 MeV//c^2.`


`color{blue} ✍️`The force that determines the motion of atomic electrons is the familiar Coulomb force. We have seen that for average mass nuclei the binding energy per nucleon is approximately `8 MeV,` which is much larger than the binding energy in atoms.

`color{blue} ✍️`Therefore, to bind a nucleus together there must be a strong attractive force of a totally different kind. It must be strong enough to overcome the repulsion between the (positively charged) protons and to bind both protons and neutrons into the tiny nuclear volume.

`color{blue} ✍️`We have already seen that the constancy of binding energy per nucleon can be understood in terms of its short-range. Many features of the nuclear binding force are summarised below. These are obtained from a variety of experiments carried out during 1930 to 1950.

`color{blue} {(i)}` The nuclear force is much stronger than the Coulomb force acting between charges or the gravitational forces between masses. The nuclear binding force has to dominate over the Coulomb repulsive force between protons inside the nucleus. This happens only because the nuclear force is much stronger than the coulomb force. The gravitational force is much weaker than even Coulomb force.

`color{blue} {(ii)}` The nuclear force between two nucleons falls rapidly to zero as their distance is more than a few femtometres. This leads to saturation of forces in a medium or a large-sized nucleus, which is the reason for the constancy of the binding energy per nucleon.

`color{blue} ✍️`A rough plot of the potential energy between two nucleons as a function of distance is shown in the Fig. 13.2. The potential energy is a minimum at a distance `r_0` of about 0.8 fm. This means that the force is attractive for distances larger than 0.8 fm and repulsive if they are separated by distances less than 0.8 fm.

`color{blue} {(iii)}` The nuclear force between neutron-neutron, proton-neutron and proton-proton is approximately the same. The nuclear force does not depend on the electric charge.

`color{blue} ✍️`Unlike Coulomb’s law or the Newton’s law of gravitation there is no simple mathematical form of the nuclear force.


`color{blue} ✍️`Becquerel discovered radioactivity in 1896 purely by accident. While studying the fluorescence and phosphorescence of compounds irradiated with visible light, Becquerel observed an interesting phenomenon.

`color{blue} ✍️`After illuminating some pieces of uranium-potassium sulphate with visible light, he wrapped them in black paper and separated the package from a photographic plate by a piece of silver.

`color{blue} ✍️`When, after several hours of exposure, the photographic plate was developed, it showed blackening due to something that must have been emitted by the compound and was able to penetrate both black paper and the silver.

`color{blue} ✍️`Experiments performed subsequently showed that radioactivity was a nuclear phenomenon in which an unstable nucleus undergoes a decay. This is referred to as radioactive decay. Three types of radioactive decay occur in nature :

`color{blue} {(i)}` `alpha`-decay in which a helium nucleus `text()_(2)^(4)He` is emitted;

`color{blue} {(ii)}` `beta`-decay in which electrons or positrons (particles with the same mass as electrons, but with a charge exactly opposite to that of electron) are emitted;

`color{blue} {(iii)}` `gamma`-decay in which high energy (hundreds of keV or more) photons are emitted.
Each of these decay will be considered in subsequent sub-sections.

`color{brown}bbul("Law of radioactive decay")`
`color{blue} ✍️`In any radioactive sample, which undergoes `alpha, beta` or g-decay, it is found that the number of nuclei undergoing the decay per unit time is proportional to the total number of nuclei in the sample. If N is the number of nuclei in the sample and `DeltaN` undergo decay in time `Deltat` then

`color{blue} ✍️` `color{green}{(DeltaN)/(Deltat) proptoN}`

`color{navy}(DeltaN//Deltat = λN,)`


`color{blue} ✍️`where `I` is called the radioactive decay constant or disintegration constant. The change in the number of nuclei in the sample* is `dN = – DeltaN` in time `Deltat`.

`color{blue} ✍️`Thus the rate of change of N is (in the limit `Deltat -> 0`)

`color{purple}((Dn)/(dt) = - lamdaN)`
`color{purple}((dN)/N =- lamdadt)`

`color{blue} ✍️`Now, integrating both sides of the above equation,we get,

`color{navy}(int_(N0)^(N) (dN)/N = - lamda int_(t_0)^(t) dt)`



`color{navy}(ln N − ln N_0 = −lamda(t – t_0))`


`color{blue} ✍️`Here `N_0` is the number of radioactive nuclei in the sample at some arbitrary time `t_0` and N is the number of radioactive nuclei at any subsequent time t. Setting `t_0 = 0` and rearranging Eq. (13.12) gives us

`color{navy}(In (N)/(N_0) = - lamdat)`


which gives

`color{navy}(N(t) = N_0 e^(−lamda t))`

....... (13.14)

`color{brown} {"Note,"}` for example, the light bulbs follow no such exponential decay law. If we test 1000 bulbs for their life (time span before they burn out or fuse), we expect that they will ‘decay’ (that is, burn out) at more or less the same time.

`color{blue} ✍️`The decay of radionuclides follows quite a different law, the law of radioactive decay represented by Eq. (13.14).

`color{blue} ✍️`The total decay rate `R` of a sample is the number of nuclei disintegrating per unit time. Suppose in a time interval dt, the decay count measured is `DeltaN.` Then `dN = – DeltaN.`

`color{blue} ✍️`The positive quantity `R` is then defined as

`color{purple}(R = - (dN)/(dt))`

`color{blue} ✍️`Differentiating Eq. (13.14), we get

`color{purple}(R = lamdaN_0e^(-lamdat))`
`color{purple}(R= R_0e^(-lamdat))`

`color{blue} ✍️`This is equivalant to the law of radioactivity decay, since you can integrate Eq. (13.15) to get back Eq. (13.14). Clearly, `R_0 = lamdaN_0` is the decay rate at `t = 0`.

`color{blue} ✍️`The decay rate `R` at a certain time t and the number of undecayed nuclei N at the same time are related by

`color{navy}(R = lamdaN)`


`color{blue} ✍️`The decay rate of a sample, rather than the number of radioactive nuclei, is a more direct experimentally measurable quantity and is given a specific name: activity. The SI unit for activity is becquerel, named after the discoverer of radioactivity, Henry Becquerel.

`color{blue} ✍️`1 becquerel is simply equal to 1 disintegration or decay per second. There is also another unit named “curie” that is widely used and is related to the SI unit as:

`color{purple}(1 "curie" = 1 Ci = 3.7 × 10^(10) "decays per second")`

`color{purple}(= 3.7 × 10^(10))` Bq

`color{blue} ✍️`Different radionuclides differ greatly in their rate of decay. A common way to characterize this feature is through the notion of half-life.
Half-life of a radionuclide (denoted by `T_(1//2)`) is the time it takes for a sample that has initially, say `N_0` radionuclei to reduce to
`color{purple}(N_0//2).` Putting `color{purple}(N = N_0//2)` and `color{purple}(t = T_(1//2))` in Eq. (13.14), we get

`color{navy}(T_(1//2)= (In2)/(lamda) = (0.693)/(lamda))`

....... (13.17)

`color{blue} ✍️`Clearly if `N_0` reduces to half its value in time `color{green}(T_(1//2), R_0)` will also reduce to half its value in the same time according to Eq. (13.16).

`color{blue} ✍️`Another related measure is the average or mean life t. This again can be obtained from Eq. (13.14).

`color{blue} ✍️`The number of nuclei which decay in the time interval t to `t + Deltat` is `R(t)Deltat = lamdaN_0e^(–lamdat)Deltat`.
Each of them has lived for time t.

`color{blue} ✍️`Thus the total life of all these nuclei would be `tlamdaN_0 e^(-lamdat) Deltat` It is clear that some nuclei may live for a short time while others may live longer. Therefore to obtain the mean life, we have to sum (or integrate) this expression over all times from `0` to `infty` , and divide by the total number `N_0` of nuclei at `t = 0.`

`color{blue} ✍️`Thus, `color{purple}(τ = (lamdaN_0 int_(0)^(infty) te^(-lamdat)dt)/(N_0) = lamdaint_(0)^(infty)te^(-lamdat)dt)`
One can show by performing this integral that

`color{purple}(τ = 1//λ)`

`color{blue} ✍️`We summarise these results with the following:

`color{navy}{T_(1//2)= (In2)/(lamda) = τIn2}`

............ (13.18)

`color{blue} ✍️`Those radioactive elements whose half-life is short compared to the age of the universe (13.7 billion years) are not found in observable quantities in nature today.

`color{blue} ✍️`They have, however, been seen in the laboratory in nuclear reactions. Tritium and plutonium belong to this category.

Q 3109678518

The half-life of `text()_(92)^(238)U` undergoing α-decay is `4.5 ×10^3` years. What is the activity of 1g sample of `text()_(92)^(238)U`
Class 12 Chapter 13 Example 4

`T_(1//2) = 4.5 × 10^9 y`
`= 4.5 × 10^9 y x 3.16 x 10^7 s//y`
`= 1.42 × 10^17s`
One k mol of any isotope contains Avogadro’s number of atoms, and so 1g of `text()_(92)^(238)U` contains
`1/(238xx10^(-3)) kmol 238 ×10× 6.025 × 10^(26) atoms//kmol`
`= 25.3 × 10^(20) atoms.`
The decay rate R is
R = λN
`= (0.693)/(T_(1//2)N) = (0.693xx25.3xx10^(20))/(1.42xx10^(17)) s^(-1)`
`= 1.23xx10^4 s^(-1)`
`= 1.23xx10^4Bq`
Q 3119678519

Tritium has a half-life of 12.5 y undergoing beta decay. What fraction of a sample of pure tritium will remain undecayed after 25 y.
Class 12 Chapter 13 Example 5

By definition of half-life, half of the initial sample will remain undecayed after 12.5 y. In the next 12.5 y, one-half of these nuclei would have decayed. Hence, one fourth of the sample of the initial pure tritium will remain undecayed.

` ` decay

`color{blue} ✍️` When a nucleus undergoes alpha-decay, it transforms to a different nucleus by emitting an alpha-particle (a helium nucleus,

`color{purple}(text()_(2)^(4) He)` For example, when Uundergoes alpha-decay, it transforms to `text()_(90)^(234)Th`



`color{blue} ✍️`In this process, it is observed that since `color{navy}(text()_(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

`color{purple}(text()_(Z)^(A)X)` into a nucleus `color{purple}(text()_(Z-2)^(A-4)Y)` can be expressed as

`color{navy}(text()_(Z)^(A)X -> text()_(Z-2)^(A-4)Y) + text()_(2)^(4)He`

...... (13.20)

`color{blue} ✍️`where `color{purple}(text()_(A)^(Z)X)` is the parent nucleus and `color{purple}(text()_(Z-2)^(A-4)Y)` is the daughter nucleus.

`color{blue} ✍️`The alpha-decay of `color{purple}(text()_(92)^(238) U)` can occur spontaneously (without an external source of energy) because the total mass of the decay products `color{purple}(text()_(92)^(238)Th)` and `color{purple (text()_(2)^(4)He)` is less than the mass of the original `color{purple}(text()_(92)^(238) U).`

`color{blue} ✍️`Thus, the total mass energy of the decay products is less than the mass energy of the original nuclide. 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

`color{navy}(Q = (m_X – m_Y – m_(He)) c^2)`


`color{blue} ✍️`This energy is shared by the daughter nucleus `color{aqua}(text()_(A-4)^(Z-2)Y)` and the alpha `color{green}(text()_(2)^(4)He)` in the form of kinetic energy. Alpha-decay obeys the radioactive law as given in Eqs. (13.14) and (13.15).

Beta decay

`color{blue} ✍️`A nucleus that decays spontaneously by emitting an electron or a positron is said to undergo beta decay. Like alpha decay, this is a spontaneous process, with a definite disintegration energy and half-life. Again like alpha decay, beta decay is a statistical process governed by Eqs. (13.14) and (13.15).

`color{blue} ✍️`In beta minus (β − ) decay, an electron is emitted by the nucleus, as in the decay of `text()_(32)^(15)P`

`color{navy}(text()_(32)^(15)P -> text()_(32)^(16)S + (e^-) + barv }`


`color{blue} ✍️`In beta plus (β+ ) decay, a positron is emitted by the nucleus, as in the decay of `color{purple}(text()_(11)^(22)Na)`

`color{blue}(text()_(11)^(22)Na -> text()_(10)^(22)Ne +e^+ + v}`

........... (13.23)

`color{blue} ✍️`The symbols `barν` and ν represent antineutrino and neutrino, respectively; both are neutral particles, with very little or no mass.

`color{blue} ✍️`These particles are emitted from the nucleus along with the electron or positron during the decay process. Neutrinos interact only very weakly with matter; they can even penetrate the earth without being absorbed. It is for this reason that their detection is extremely difficult and their presence went unnoticed for long.

`color{blue} ✍️`In beta-minus decay, a neutron transforms into a proton within the nucleus according to

`color{navy}(n → p + e^– + barν)`


`color{blue} ✍️`whereas in beta-plus decay, a proton transforms into neutron (inside the nucleus) via

`color{navy}(p → n + e^+ + ν)`


`color{blue} ✍️`These processes show why the mass number A of a nuclide undergoing beta decay does not change; one of its constituent nucleons simply changes its character according to Eq. (13.24) or (13.25).

Gamma decay

`color{blue} ✍️`Like an atom, a nucleus also has discrete energy levels - the ground state and excited states. The scale of energy is, however, very different. Atomic energy level spacings are of the order of eV, while the difference in nuclear energy levels is of the order of MeV.

`color{blue} ✍️`When a nucleus in an excited state spontaneously decays to its ground state (or to a lower energy state), a photon is emitted with energy equal to the difference in the two energy levels of the nucleus. This is the so-called `"gamma decay."`

`color{blue} ✍️`The energy (MeV) corresponds to radiation of extremely short wavelength, shorter than the hard X-ray region.

`color{blue} ✍️`Typically, a gamma ray is emitted when a α or β decay results in a daughter nucleus in an excited state. This then returns to the ground state by a single photon transition or successive transitions involving more than one photon. A familiar example is the successive emmission of gamma rays of energies 1.17 MeV and 1.33 MeV from the deexcitation of `text()_60^28Ni` nuclei formed from `β^-`
decay of `text()_60^27Co`

`color{blue} ✍️`This process is depicted in Fig.13.4 through an energy level diagram.