
1.Basics
10
Lecture1.1

Lecture1.2

Lecture1.3

Lecture1.4

Lecture1.5

Lecture1.6

Lecture1.7

Lecture1.8

Lecture1.9

Lecture1.10


2.Electrostatics (1)
8
Lecture2.1

Lecture2.2

Lecture2.3

Lecture2.4

Lecture2.5

Lecture2.6

Lecture2.7

Lecture2.8


3.Electrostatics (2)
7
Lecture3.1

Lecture3.2

Lecture3.3

Lecture3.4

Lecture3.5

Lecture3.6

Lecture3.7


4.Current Electricity (1)
9
Lecture4.1

Lecture4.2

Lecture4.3

Lecture4.4

Lecture4.5

Lecture4.6

Lecture4.7

Lecture4.8

Lecture4.9


5.Current Electricity (2)
4
Lecture5.1

Lecture5.2

Lecture5.3

Lecture5.4


6.Capacitor
6
Lecture6.1

Lecture6.2

Lecture6.3

Lecture6.4

Lecture6.5

Lecture6.6


7.RC Circuits
3
Lecture7.1

Lecture7.2

Lecture7.3


8.Magnetism and Moving Charge
16
Lecture8.1

Lecture8.2

Lecture8.3

Lecture8.4

Lecture8.5

Lecture8.6

Lecture8.7

Lecture8.8

Lecture8.9

Lecture8.10

Lecture8.11

Lecture8.12

Lecture8.13

Lecture8.14

Lecture8.15

Lecture8.16


9.Magnetism and Matter
10
Lecture9.1

Lecture9.2

Lecture9.3

Lecture9.4

Lecture9.5

Lecture9.6

Lecture9.7

Lecture9.8

Lecture9.9

Lecture9.10


10.Electromagnetic Induction
14
Lecture10.1

Lecture10.2

Lecture10.3

Lecture10.4

Lecture10.5

Lecture10.6

Lecture10.7

Lecture10.8

Lecture10.9

Lecture10.10

Lecture10.11

Lecture10.12

Lecture10.13

Lecture10.14


11.Alternating Current Circuit
8
Lecture11.1

Lecture11.2

Lecture11.3

Lecture11.4

Lecture11.5

Lecture11.6

Lecture11.7

Lecture11.8


12.Electromagnetic Waves
4
Lecture12.1

Lecture12.2

Lecture12.3

Lecture12.4


13.Photoelectric Effect
5
Lecture13.1

Lecture13.2

Lecture13.3

Lecture13.4

Lecture13.5


14.Ray Optics (Part 1)
12
Lecture14.1

Lecture14.2

Lecture14.3

Lecture14.4

Lecture14.5

Lecture14.6

Lecture14.7

Lecture14.8

Lecture14.9

Lecture14.10

Lecture14.11

Lecture14.12


15.Ray Optics (Part 2)
14
Lecture15.1

Lecture15.2

Lecture15.3

Lecture15.4

Lecture15.5

Lecture15.6

Lecture15.7

Lecture15.8

Lecture15.9

Lecture15.10

Lecture15.11

Lecture15.12

Lecture15.13

Lecture15.14


16.Ray Optics (Part 3)
6
Lecture16.1

Lecture16.2

Lecture16.3

Lecture16.4

Lecture16.5

Lecture16.6


17.Wave Optics
21
Lecture17.1

Lecture17.2

Lecture17.3

Lecture17.4

Lecture17.5

Lecture17.6

Lecture17.7

Lecture17.8

Lecture17.9

Lecture17.10

Lecture17.11

Lecture17.12

Lecture17.13

Lecture17.14

Lecture17.15

Lecture17.16

Lecture17.17

Lecture17.18

Lecture17.19

Lecture17.20

Lecture17.21


18.Atomic Structure
6
Lecture18.1

Lecture18.2

Lecture18.3

Lecture18.4

Lecture18.5

Lecture18.6


19.Nucleus
6
Lecture19.1

Lecture19.2

Lecture19.3

Lecture19.4

Lecture19.5

Lecture19.6


20.XRay
4
Lecture20.1

Lecture20.2

Lecture20.3

Lecture20.4


21.Error and Measurement
9
Lecture21.1

Lecture21.2

Lecture21.3

Lecture21.4

Lecture21.5

Lecture21.6

Lecture21.7

Lecture21.8

Lecture21.9


22.Semiconductors
9
Lecture22.1

Lecture22.2

Lecture22.3

Lecture22.4

Lecture22.5

Lecture22.6

Lecture22.7

Lecture22.8

Lecture22.9


23.Communication Systems
5
Lecture23.1

Lecture23.2

Lecture23.3

Lecture23.4

Lecture23.5

Chapter Notes – Nucleus
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 N_{o} 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 halflife, T, and may be related to λ as follows. Since
0.5N0=N0e−λt
we have λT=ln2=0.693. Therefore
T=0.693λ
It takes one halflife to drop to 50% of any starting value. The halflife for the decay of the free neutron is 12.8 min. Other halflives range from about 10^{20} s to 10^{16 }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 10^{10} dps
1 rutherford = 10^{6} 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 halflife of Cobalt – 60 is 5.25 years. How long after its activity have decreased to about oneeight of its original value ?
Solution:
The activity is proportional to the number of undecayed atoms.
In each halflife, half the remaining sample decays.
Since (12)(12)(12)=18 , therefore, three halflives or 15.75 years are required for the sample to decay to 18th its original strength.
Example 2
A count ratemeter 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 halflife 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 aparticles is relatively low: v_{a} = (c/30 to c /15), where c is the velocity of light.
(2) β–decay (betaminusdecay) associated with the emission of electrons formed at the instant of decay.
Both processes are accompanied by γradiation, i.e. the flow of photons having a very small wavelength, and hence a very high energy. Like other electromagnetic waves, γrays propagate at a velocity of light. The penetrability of γrays is 1100 times higher than the penetrability of βrays and 100010000 times higher than the penetrability of αrays. It also exceeds the penetrability of Xrays.
In a magnetic field, a beam of α, β, and γrays splits into three parts as shown in the figure.
Nuclei possessing the artificial radioactivity are obtained by bombarding stable nuclei of heavy elements by αparticles, 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 αdecay, 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 βdecay, 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 βradiation. 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 _{6}C^{12} 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 C^{12} has a mass of 12 g and contains Avogadro’s number, N_{A}, of atoms.
By definition, each C^{12} has a mass of 12 u.
Thus, 12 g corresponds to 12 N_{A} 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 _{8}O^{16}.
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 10^{14} times the density of water.