Class 12 Physics Chapter 13 Nuclei Notes

Nuclei

Radioactivity

Radioactive decay is a random process. Each decay is an independent event, and one cannot tell when a particular nucleus will decay. When a given nucleus decays, it is transformed another nuclide, which may or may not be radioactive. When there is a very large number of nuclei in a sample, the rate of decay is proportional to the number of nuclei N that are present

dNdt=−λN

Where λ is called the decay constant. This equation may be expressed in the form dNN=−λdt and integrated

∫NoNdNN=−λ∫0tdt
to yield
ln(NNo)=−λt

where No is the initial number of parent nuclei at t = 0. The number that survive at time t is therefore

N=N0e−λt

This function is plotted in the following figure.The time required for the number of parent nuclei to fall to 50% is called the half-life, T, and may be related to λ as follows. Since

0.5N0=N0e−λt

we have λT=ln|2|=0.693. Therefore

T=0.693λ

It takes one half-life to drop to 50% of any starting value. The half-life for the decay of the free neutron is 12.8 min. Other half-lives range from about 10-20 s to 1016 years.
Since the number of atoms is not directly measurable, we measure the decay rate or activity (A)
A=−dNdt. On taking the derivative of equation we find          A=λN=A0e−λt
where A0=λN0 is the initial activity. The SI unit for the activity is the becquerel (Bq), but the curie (Ci) is often used in practice.

1 becquerel (Bq)                =          1 disintegration per second (dps)
1 curie                              =          3.7 x 1010 dps
1 rutherford                      =           106 dps
Mean life of a radioactive sample is defined as the average of the lives of all nuclei.

i.e.  Tav=∫0∞Noe−λtdtNo=1λ=T0.693

Example 1
The half-life of Cobalt - 60 is 5.25 years. How long after its activity have decreased to about one-eight of its original value ?

Solution:

The activity is proportional to the number of undecayed atoms.
In each half-life, half the remaining sample decays.
Since (12)(12)(12)=18  , therefore, three half-lives or 15.75 years are required for the sample to decay to 18th its  original strength.

Example 2
A count rate-meter is used to measure the activity of a given sample. At one instant the meter shows 4750 counts per minute. Five minutes later it shows 2700 counts per minute.
(a)  Find the decay constant
(b)  Also, find the half life of the sample

Solution:

Initial velocity Ai=dNdt∣∣t=0=λNo=4750       (i)
Final velocity Af=dNdt∣∣t=5=λN=2700        (ii)
Dividing (i) by (ii), we get
47502700=NoNt    (iii)
The decay constant is given by

λ=2.303tlogNoNt    or       λ=2.303tlog47502700=0.113min−1

Half life of the sample is

T=0.693λ=0.6930.113=6.14min

Example 3
The mean lives of a radio active substance are 1600 and 400 years for α - emission and β - emission respectively. Find out the time during which three fourth of a sample will decay if it is decaying both by α - emission and β - emission simultaneously.

Solution:

When an substance decays by a and b emission simultaneously, the average disintegration constant λav is given by

λav=λα+λβ

where λα = disintegration constant for α - emission only
          λβ = disintegration constant for β - emission only
Mean life is given by   Tm=1/λav
⇒λav=λα+λβ   or   1Tm=1Tα+1Tβ=11600+1400=3.12×10−3
λavt=2.303logN0Nt
⇒(3.12×10−3)t=2.303log10025
⇒t=2.303×13.12×10−3log4=443.5years

Example 4
The half-life of radium is 1620 years. How many radium atoms decay in 1s in a 1g sample of radium. The atomic weight of radium is 226 g/mol.

Solution:

Number of atoms in 1 g sample is

N=(0.001226)(6.02×1026)=2.66×1021atoms

The decay constant is

λ=0.693T1/2=0.693(1620)(3.16×107)=1.35×10−11s−1

Taking one year =3.16×107s
Now,   ΔNΔt=λN=(1.35×10−11)(2.66×1021)=3.6×1010s−1
Thus    3.6×1010 nuclei decay in one second.

Properties of Radioactive Processes

(1)  α - decay associated with the emission of α - particles, viz. nuclei 42Heof helium. Alpha particles are heavy positively charged particles having a mass mα≈4 amu  and a charge qα=+2e. The velocity of a-particles is relatively low: va = (c/30 to c /15), where c is the velocity of light.
(2)  β-decay (beta-minus-decay) associated with the emission of electrons formed at the instant of decay.

Both processes are accompanied by γ-radiation, i.e. the flow of photons having a very small wavelength, and hence a very high energy. Like other electromagnetic waves, γ-rays propagate at a velocity of light. The penetrability of γ-rays is 1-100 times higher than the penetrability of β-rays and 1000-10000 times higher than the penetrability of α-rays. It also exceeds the penetrability of X-rays.
In a magnetic field, a beam of α-, β-, and γ-rays splits into three parts as shown in the figure.

Nuclei possessing the artificial radioactivity are obtained by bombarding stable nuclei of heavy elements by α-particles, neutrons, or (sometimes) protons and other particles. Nuclear transformations occur in two stages in this case. First a particle hits a target nucleus and causes its transformation into another, unstable (radioactive), nucleus. This newly formed nucleus spontaneously emits a particle and is transformed either into a stable nucleus or into a new radioactive nucleus. Artificial radioactivity obeys the same laws as natural radioactivity.

Radioactive processes occur in accordance with the laws of conservation of energy, momentum, angular momentum, electric charge, and mass number (amount of nucleons).
In α-decay, the mass number of the nucleus decreases by four and the charge decreases by two units, as a result of which two electrons are removed from the atomic shell. The element transforms into another element with the atomic number which is two units lower.
In β-decay, a neutron in the nucleus transforms into a proton. Such a transformation of the neutral neutron into the positive proton is accompanied by the birth of an electron, i.e. by β-radiation. The mass number of the nucleus does not change in this process, while the charge increases by +e and atomic number increases by one.

ATOMIC NUCLEUS

The atomic nucleus consists of two types of elementary particles, viz. protons and neutrons. These particles are called nucleons.
The proton (denoted by p) has a charge +e and a mass mp≈1.6726×10−27kg, which is approximately 1840 times larger than the electron mass. The proton is the nucleus of the simplest atom with Z = 1, viz. the hydrogen atom.

The neutron (denoted by n) is an electrically neutral particle (its charge is zero). The neutron mass mn≈1.6749×10−27kg. The fact that the neutron mass exceeds the proton mass by about 2.5 electron masses is of essential importance. It follows from this that the neutron in free state (outside the nucleus) is unstable (radioactive). During the time equal on the average to 12 min, the neutron spontaneously transforms to the proton by emitting an electron (e-) and a particle called the antineutrino (v~). This process can be schematically written as follows:

n→p+e−+v~

The most important characteristics of the nucleus are the charge number Z (coinciding with the atomic number of the element) and the mass number A. The charge number Z is equal to the number of protons in the nucleus, and hence it determines the nuclear charge equal to Ze. The mass number A is equal to the number of nucleons in the nucleus (i.e. to the total number of protons and neutrons).
Nuclei are symbolically designated as

XAZ  or   ZXA

where X stands for the symbol of a chemical element. For example, the nucleus of the oxygen atom is symbolically written as O188 or 8O18.
Most of the chemical element have several types of atoms differing in the number of neutrons in their nuclei. These varieties are called isotopes. For example, oxygen has three stable isotopes: O168, O178 and O188. In addition to stable isotopes, there also exist unstable (radioactive) isotopes.
Atomic masses are specified in terms of the atomic mass unit or unified mass unit (u). The mass of a neutral atom of the carbon isotope 6C12 is defined to be exactly 12 u.

1u=1.66056×10−27kg=931.5MeV

Example 5
(a)  Calculate the value of 1 u from Avogadro’s number.
(b)  Determine the energy equivalent of 1u.

Solution:

(a)  One mole of C12 has a mass of 12 g and contains Avogadro’s number, NA, of atoms.
By definition, each C12 has a mass of 12 u.
Thus, 12 g corresponds to 12 NA u   which means
1u=1gNA=16.022045×1023
or   1u=1.66056×10−27kg

(b)  From Einstein relation  E=mc2
⸫      E=(1.66056×10−27)(3×108)2=1.4924×10−10J
Since   1eV=1.6×10−19J
⸫         E=931.5MeV
Hence 1u=931.5MeV

Size of the Nucleus

The shape of nucleus is approximately spherical and its radius is approximately related to the mass number by
R≈1.2A1/3fm         where  1 fermi (fm) = 10-15 m 

Example 6
Find the mass density of the oxygen nucleus 8O16.

Solution:

Volume     V=43πR3=43π(1.2)3A=1.16×10−43m3
Mass of oxygen atoms (A = 16) is approximately 16 u.
Therefore, density is ρ=mv
or         ρ=(16)(1.66×10−27)1.16×10−43=2.3×1017kg/m3
This is 1014 times the density of water.