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CHAPTER – 5
MODERN PHYSICS
NUCLEAR STRUCTURE
(a) CompositionThe central core of an atom is called as nucleus. The nucleus of any atom of an element hastwo types of fundamental particles, i.e., protons and neutrons. Collectively all the particlesfound within the nucleus are called nucleons.
(b) ChargeNucleus owes its positive charge due to protons. Among all the elements, the nuclear chargeis minimum for the hydrogen and it is equal to the charge of 1 proton. Naturally occurringelement whose nuclear charge is maximum is uranium and it is equal to the charge of 92protons.
The proton is a positively charged particle having a charge of + 1.6 10-19 coulombs andmass kg. However, its mass is 1837 times greater than the mass of oneelectron.The neutron apparently has no electric charge, and its mass is slightly greater than the massof one proton.
(c) Representation of NucleusMass number (A): It represents total number of nucleons present within the nucleus, i.e.,the number of protons and neutrons. It is also known as mass number or atomic weight.
Atomic number (Z): It represents total number of protons, within the nucleus of an atom. Itis numerically equal to the number of electrons revolving around the nucleus. It is also
known as atomic number or proton number. It controls the chemical and physicalproperties of an element.N represents total number of neutrons present in the nucleus of an atom.
Thus,Mass number = Number of protons + Number of neutrons or A = Z + N
If X is the symbol of some element, such that A is its mass number and Z is its atomicnumber then symbolically its nucleus is represented as ZX A For example, for oxygen A = 16 and Z = 8 and hence symbol for its nucleus is 8O16 It must be kept in mind that the superscript represents mass number and sub-script
represents atomic number.
(d) Mass of the nucleusIt is expressed in terms of the unit called atomic mass unit (a.m.u)
Atomic mass unit: the mass of molecules, atoms, nucleus and atomic particles is measuredin a unit called “atomic mass unit”. This is denoted by „a.m.u‟. One atomic mass unit is one-twelfth of the mass of a carbon
atom.1 a.m.u = 1.66 × 10-24 gm
(e) Size of nucleusRadius of nucleus is used to measure the size of the nucleus. The unit used for measuringthe size of the nucleus is Fermi (F). 1 Fermi = mThe radius of a nucleus is,
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Where R is radius of nucleus, A mass number and R o is a constant. R o = 1.2 Fermi
(f) Volume of nucleusSince nucleus is considered to be spherical, its volume could be found by using the formulafor the volume of the sphere.
Therefore, volume of the nucleus = V =
V=
V=
V Volume of nucleus is directly proportional to its mass number (A).
(g) Density of nucleus As we know, density = mass/volumeMass of nucleus = total no. of nucleons * mass of one nucleon
=A×1 a.m.u
Volume of nucleus =
Density of nucleus =
Density of nucleus is a constant number; therefore we can say that it is independent of theelement.
Classification of atoms based on nucleons
1. IsotopesIsotopes are the atoms of same element having same atomic number, but different atomicmasses due to difference in number of neutrons in their nucleus.
Example of Isotopes:a. Hydrogen has three isotopes,
1H1 Hydrogen (Protium), 1H2 Deuterium and 1H3 Tritium. b. Chlorine has two isotopes, i.e., 17Cl35 and 17Cl37.c. Oxygen has three isotopes, i.e., 8O16, 8O17 and 8O18 d. Uranium has two isotopes, i.e., 92U235 and 92U238.e. Carbon has 3 isotopes, i.e., 6C12, 6C13 and 6C14
Here, it must be pointed out that practically every element has its isotopes. There areabout one thousand unstable isotopes of radioactive elements and 300 isotopes of stableelements. Isotopes differs in their mass number, therefore, isotopes differ in physicalproperties.
2. Isobars Atoms of different elements, having same mass number, but different number of protons(atomic number) are called isobars. Isobars differ in number of electrons, protons andneutrons.
An isobars differs in atomic number as well as number of protons, they differ in physical as well as in chemical properties.
Examples of Isobarsa. 18 Ar40 and 20Ca40 are isobars
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b. 12Mg24 and 11Na24 are isobars.
3. Isotones Atoms of different elements with same number of neutrons are called isotones. They differin number of electrons and protons.
Examples of isotones
14Si30 and 15P31
Forces inside the nucleus
There are two major forces present inside the nucleusi. Electrostatic force: this is repulsive force between the protons. This force depends on
the charge on the nucleons. Electrostatic force impact is felt at larger distance.ii. Nuclear force: this is an attractive force among the nucleons. The impact of nuclear
force is felt only at very small distances within the nucleus. Hence nuclear force is calledshort range force. Nuclear force is independent of charge of nucleons.
As the nuclear force is stronger than electrostatic force the nucleons do not fly apart.
Mass defect & Binding energy
The difference between the expected mass and experimentally measured mass of nucleus iscalled mass defect. It is denoted by Δm. It can be calculated as follows: Mass defect (Δm) = (expected mass of nucleus) - (measured mass of nucleus)
Δm = (sum of masses of protons and neutrons) - (measured mass of nucleus)
When the nucleons are grouped together to form a nucleus, they lose a small amount of mass,i.e., there is mass defect. This mass defect is released as radiant energy according to therelation E = mc2;Thus
Energy released = mass defect × c2.This energy is used to bind nucleons together in the nucleus; therefore it is called bindingenergy.The smallest unit of binding energy for one a.m.u is generally mentioned in terms of MeV.
Thus energy equivalent of 1 a.m.u of mass is 931.5 MeV.
If mass effect (Δm) is expressed in terms of a.m.u, then binding energy = Δm931.5 MeV
In nuclear reactions, the energy that must be radiated or otherwise removed as binding energy may be in the form of electromagnetic waves, such as gamma radiation, or as heat.
Average Binding energy
Average binding energy is the energy required to remove one nucleon from the nucleus.Binding energy per nucleon = Binding energy of the nucleus/ total number of nucleons
The stability of nucleus is understood in terms of binding energy per nucleons. More is the
binding energy per nucleon, more is the stability.
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Graph of Binding energy per nucleon and mass number A
The main observations from the graph are:a. The binding energy per nucleon is smaller for lighter nuclei and increases with mass
number b. The nuclei in the mass number range 40 to 120 are most stable as the binding energy
per nucleon is more in their case. The peak value appears around A =60. The maximum binding energy per nucleon is for iron with a value of 8.7 MeV/nucleon.
c. For nuclei with mass number higher than 120, the binding energy per nucleon starts
decreasing. The nuclei below 7.6 MeV/ nucleon like 92U238
are unstable.
Stability of Nuclei
There are several factors affecting nuclear stability, many of which are reflected in the abundance curve:
a. As number of nucleon increases, attraction between nucleons increases also value of binding
energy increase, therefore stability of nuclei increases.
b. Magic numbers: Nuclei which have 2,8,20,28,50,82 and 126 number of neutrons or protons are
more abundant in nature than other nuclei. This suggests that these nuclei are more stable than
others. The above sequence of numbers is called Magic numbers.
c. Surface tension (unrequited bonds): smaller nuclei are less stable because they have Highsurface/volume ratios.
d. Coulomb or electrostatic repulsion works against larger nuclei, because you pay a price stuffing
charge into a small volume. Although you always gain by adding more nucleons, but at large
distance nuclear force become weaker in comparison to electrostatic force
Just as every phenomenon tries to achieve stability, unstable nuclei try to achieve stability by throwing
out excess particles in the form of radiations. This gave rise to a phenomenon which was first observed
by Henry Becquerel. This phenomenon was called radioactivity by Madam Curie.
Radioactivity
The spontaneous disintegration of the nuclei with the emission of certain particles andradiations is called radioactivity. It is the phenomenon due to which certain elements given out
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highly penetrating radiations spontaneously. The activity of a radioactive substance is definedas the number of atoms disintegrating in one second. It is measured in various units. The SIunit of radioactivity is Becquerel.
a. 1 Becquerel (1 Bq) = 1 disintegration per second. b. 1 Curie (1 Ci) = disintegration per second = activity of 1 gm of pure radium
Causes of Radioactivity As the atomic number and mass number increases, the size of the nucleus also increases.The increase in size leads to weakening of nuclear force and the repulsive force startsplaying a prominent role. In the element with atomic number Z=83 (Bismuth) the repulsiveforce just exceeds the attractive nuclear force, making the nucleus unstable. Emission startsoccurring from such unstable nucleus and then tries to attain stability by emitting particlesfrom inside it. This process of emission is known as radioactivity.
Radioactive elementsThe elements which give out high energy radiations on their own are called radioactive
elements.
Examples of Radioactive elements :1. Uranium 2. Thorium 3. Radium 4. Polonium.
Becquerel RaysThe original name for the radiations give out by radioactive elements is Becquerel rays.Properties of Becquerel rays:
a. They affect photographic plate. b. They ionize the gas through which they pass.c. They can penetrate through matter. The range of penetration depends upon nature
of radioactive source and density of matter.d. They are affected by electrostatic and magnetic fields.
Radioactive emission has three types of radiations – α, β and γ radiations
Comparative Table of α, β and γ Radiations
Properties -particles(2He4) -particles(-1e0) -rays
Name Are doubly chargedhelium nuclei
Electrons moving withhigh velocity
Electromagnetic waves
Rest mass 6.64 10-27 kg 9.106 10-31 kg Nil Velocity 1.4 to 2.3 107 m/s 1.1 to 2.96 108 m/s 3 108 m/sNature of charge Positive Negative Neutral
Amount of charge + 3.2 10-19 C -1.6 10-19 C Nil
Specific chargeq
m
4.8 107 C/kg 1.78 1011 C/kg Nil
Wavelength Nil Nil 10-12 m to 10-14 mEffect of magneticfield
Slightly deflected Deflected more than and opposite indirection to
Not deflected
Effect of electrostatic field
Deflected towardsnegative plate
Deflected towardspositive plate
Not deflected
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Penetrating power 2.7 cm to 8.6 cm inair
5 mm of Al or 1 mm of lead
30 cm of iron
Relative ionizingpower
10,000 100 1
Laws of Radioactive Emission
1. Rutherford and Soddy‟s law of Alpha Emission
When a radioactive nuclide ejects an alpha particle, its mass number decrease by 4 andatomic number decreases by 2 such that the position of daughter nuclide is two places
behind in the periodic table as compared to the parent nuclide.
Consider radioactive nuclide X with mass number A and atomic number Z, such that itejects an alpha particle. According to above law, the mass number of daughter nuclide will
be A – 4 and atomic number Z – 2. Representing symbolically,
ZX A Z- 2X A 4 + 2He4 Examples of -decay
92U238 90Th234 + 2He4 :
88Ra226 86Rn222 + 2He4
86Rn222 84Po218 + 2He4 :
84Po218 82Pb214 + 2He4
2. Rutherford and Soddy‟s law of Beta Emission
When a radioactive nuclide ejects beta particle, its mass number remains unaffected, butatomic number increases by one such that position of daughter nuclide is, one place aheadin the periodic table, as compared to the parent nuclei.
Beta emission generally takes place in such nuclides, which have far excess neutrons ascompared to protons. However, it is not a hard and fast rule. During beta emission, one of the neutrons breaks into an electron and proton, with release of energy. The proton remains
within the nucleus, but the electron is ejected out.
Since number of nucleons in the parent nuclide remains same as in daughter nuclide,therefore, mass number remains unchanged. However, as one proton becomes more thanparent nuclide, therefore atomic number increases by one.
0n1 1p1 + -1e0 neutron proton electron
If X is the parent nuclide of mass number A and atomic number Z, such that it ejects betaparticle, then the mass number of daughter nuclide will be A, but atomic number will
become Z+1. Representing symbolically,
ZX A Z + 1X1 A + -1e0
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Example of Beta Emission
11Na24 12Mg24 + -1e0 ;
6C14
7N14
+ -1e0
;
15P32 16S32 + -1e0
82Pb214 83Bi214 + -1e0 ;
83Bi214 84Po214 + -1e0
3. Gamma Emission
It has been found that the emission of alpha or beta particles is followed by the emission of
gamma rays. This emission takes place, because either the parent nuclide or the daughternuclide is in excited state, i.e., they have excess energy than required to hold nucleonstogether.The gamma radiation takes away neither any mass nor charge from parent nuclide, andhence the mass number and atomic number remain unchanged.
Difference between Radioactive change and Chemical change
Radioactive change Chemical change1. In radioactive change, the nucleus of
an element on disintegration
produces completely new elementsand it is essentially a nuclearphenomenon.
2. Radioactive change releases largeamount of energy.
3. Radioactive disintegration is aspontaneous event.
4. In radioactive disintegration, chargedparticles like alpha, beta are ejected.
1. Chemical change does not producenew elements and it is not related to
the nucleus of the atom.2. In a chemical change a very small
quantity of energy is released orabsorbed.
3. Chemical change is an induced event.4. In chemical change, no such ejection
of particles takes place.
Nuclear Fission
Nuclear fission is a nuclear reaction in which the nucleus of an atom splits into smaller parts(lighter nuclei), often producing free neutrons and photons (in the form of gamma rays), andreleasing a tremendous amount of energy. The two nuclei produced are most often of comparable but slightly different sizes, typically with a mass ratio of products of about 3 to 2,for common fissile isotopes. Nuclear fission was discovered by Ottohann and Strassmann in1939.
Fission is usually an energetic nuclear reaction induced by a neutron. Fission of heavy elementsis an exothermic reaction which can release large amounts of energy both as electromagnetic
radiation and as kinetic energy of the fragments (heating the bulk material where fission takesplace). In order for fission to produce energy, the total binding energy of the resulting elements
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must be greater than that of the starting element. Fission is a form of nuclear transmutation because the resulting fragments are not the same element as the original atom.
Nuclear fission produces energy for nuclear powerand to drive the explosion of nuclear weapons. Both
uses are possible because certain substances callednuclear fuels undergo fission when struck by fissionneutrons, and in turn emit neutrons when they
break apart. This makes possible a self-sustainingchain reaction that releases energy at a controlledrate in a nuclear reactor or at a very rapiduncontrolled rate in a nuclear weapon.
Most general nuclear fission reaction is given below:
92U235
+ 0n1
56Ba141
+ 36Kr92
+3 0n1
+ Q Where Q is the energy released in nuclear fission.
In the above reaction 1 neutron produces 3 neutrons, which collide with another 3 nuclei of uranium, which produces another 9 neutrons. So this reaction rapidly increases till all uraniumused. This is example of an uncontrolled chain reaction.
For controlled chain reaction, following precautions has to be taking in nuclear plants.
a. The nuclear reactors should be embedded in thick concrete walls so as to prevent any leakage of gamma radiations or neutrons.
b. The nuclear material must be kept in thick lead containers, with narrow mouth and plugged with thick lead corks.
c. Workers in nuclear establishments must wear lead lined aprons and gloves. They must wearspecial lead glasses to protect eyes.
d. Any nuclear material should be handled with mechanical tongs.e. The workers must wear special film badges. These badges can absorb nuclear radiation
which can be tested to find the amount of radiation absorbed by a particular worker.f. A periodic compulsory medical checkup should be done. If a particular person is found be
done. If a particular person is found over-exposed to the nuclear radiation, he should beimmediately withdrawn from the nuclear establishment.
Nuclear Fusion
Nuclear fusion is the process by which two or more atomic nuclei join together, or "fuse", toform a single heavier nucleus. This is usually accompanied by the release or absorption of largequantities of energy. Fusion is the process that powers active stars, the hydrogen bomb andexperimental devices examining fusion power for electrical generation.
The fusion of two nuclei with lower masses than iron (which, along with nickel, has the largest binding energy per nucleon) generally releases energy, while the fusion of nuclei heavier thaniron absorbs energy. This means that fusion generally occurs for lighter elements only, andlikewise, that fission normally occurs only for heavier elements
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Creating the required conditions for fusion on Earth is very difficult, to the point that it has not been accomplished at any scale for Protium, the common light isotope of hydrogen thatundergoes natural fusion in stars. In nuclear weapons, some of the energy released by anatomic bomb (fission bomb) is used to compress and heat a fusion fuel containing heavierisotopes of hydrogen, and also sometimes lithium, to the point of "ignition". At this point, the
energy released in the fusion reactions is enough to briefly maintain the reaction.
Fusion-based nuclear power experiments attempt tocreate similar conditions using less dramatic means,although to date these experiments have failed tomaintain conditions needed for ignition long enoughfor fusion to be a viable commercial power source.
Fusion reaction which occurs in stars and sun is following:
1H2
+ 1H3
2He4
+ 0n1
+ Q
Where Q is the energy released in single fusion.
Formulae
a. Mass of Neutron Mass of Proton
b. Number of Proton = Number of Electron = Atomic Number
c. Number of Neutron = Mass Number – Atomic Number i.e., n = A – Z
d. Charge on Electron = -1.6 10-19 Coulomb
e. Charge on Proton = + 1.6 10-19 Coulomb
f. Charge on Neutron = 0
g. Mass of Electron = 9.1 10-31 kg
h. Mass of Proton = 1.6 10-27 kg
i. 1 eV = 1.6 10-19 Joule
j. 1 a.m.u = 931 MeV
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Problems
1. What do you mean by :
(a) Atomic number(b) Mass number of a nucleus?
2. What are isobars ? Give one example.
3. What is radioactivity? Name two radioactive substances.
4. (i) What are , and -radiations composed of?
(ii) Which has least penetrating power?
5. How are -radiations produced?
6. An -particle absorbs an electron. What does is change to?
7. What type of change takes place in the nucleus when a -particle is emitted?
8. A nucleus of radioactive phosphorus has atomic number 15 and mass number 32.(i) If stable isotope of the above mentioned nucleus has one neutron less, what are the
atomic number and mass number of the isotope?(ii) If the radioactive isotope emits a -particle. What are the atomic number and mass
number of the new nucleus?
9. The element 11Na24 emits a -particle to change into Mg. Write the symbolic equation for
the change. What are the numbers 11 and 24 called ? What do they signify for the elementNa ?
10. An -particle absorbs two electrons. What does it change to ?
11. A nucleus of 88Ra226 emits an -particle. Represent the charge by an equation. Identify thenew nucleus formed, from the following :
84Po, 85 At, 86Rn, 87Fr
12. Why do isobars have different chemical properties ?
13. A radioactive element zX A loses two successive -particles and then an -particle such that
the resulting nucleus is Q Y P. Find the values of P and Q.
14. What do you understand by the term chain reaction?
15. Why is it difficult to start a fusion reaction?
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Objective Problems
1.
The mass of an electron is equal to(a) 9.1 10-31 kg (b) 9.1 10-30 kg(c) 9.1 10-27 kg (d) none of these
1. What is the particle Y in the following nuclear reaction ?
4Be9 + 2He4 6C12 + Y (a) Electron (b) Neutron (c) Proton (d) None of these
2. Nucleons correspond to the number of (a) Electrons (b) protons (c) neutrons (d) neutrons and protons
3. The mass of an atom is expressed in(a) kg (b) g (c) a.m.u. (d) Carats
4. The specific charge of an electron is equal to(a) – 1.76 1011 C kg-1 (b) 1.76 10-11 C kg-1 (c) 1.602 10-19 C (d) – 1.602 10-19 C
5. In the following nuclear reaction, how many alpha particles are given out ?
92X234 88 Y 218 (a) 2 (b) 3 (c) 4 (d) 6
6. The number of neutrons contained in 92U238 is(a) 92 (b) 146 (c) 238 (d) 330
7. In the following nuclear reaction, what is the element X ?i. 7N14 + 2He4 1H2 + X
(a) 8O16 (b) 7N14 (c) 9F17 (d) 10Ne17
8. The number of electrons contained in the nucleus of 92U235 is(a) 92 (b) 143 (c) 235 (d) zero
9. An element X with mass number A and atomic number Z is represented by (a) X (b) A XZ (c) ZX A (d) ZX A – Z
10. The process of fusion is used for constructing a(a) atom bomb (b) ordinary bomb (c) hydrogen bomb (d) neutron bomb
11. In a nuclear reactor, the moderator is(a) uranium – 234 (b) uranium – 238 (c) cadmium (d) heavy water
12. Radioactivity was discovered by (a) Rutherford (b) Becquerel (c) Bohr (d) Madame Curie
13. The discovery of neutron was made by
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(a) Chadwick (b) Rutherford (c) Becquerel (d) Curie
14. Which of the following has the least penetrating power ?(a) Alpha particles (b) Beta particles(c) Gamma rays (d) All have the same penetrating power
15. The fuel used in nuclear reactors is(a) U – 235 (b) U – 236 (c) U – 234 (d) U – 238
16. The longest and shortest wavelengths (in microns) for red and violet colours arerespectively given by (a) 0.4, 0.7 (b) 0.3, 0.6 (c) 0.7, 0.9 (d) none of these
17. The fission of uranium – 235 by means of slow moving neutrons is called a/an(a) chain reaction (b) irreversible chemical reaction(c) explosion (d) none of these
18. Which amongst the following can cause the fission of U – 235 to yield a chain reaction ?(a) Slow electrons (b) Slow protons (c) Slow neutrons (d) Fast neutrons
19. The release of energy when fission occurs in a nuclear chain reaction is the basis of (a) atom bomb (b) hydrogen bomb (c) neutron bomb (d) none of these
20. A hydrogen bomb is a(a) controlled nuclear fission (b) uncontrolled nuclear fission(c) uncontrolled nuclear fusion (d) controlled nuclear fusion
21. Which of the following will produce lesser pollution problems ?(a) Nuclear fission(b) Nuclear fusion(c) Both will produce the same amount of pollution(d) Nothing can be decided
22. Which of the following radiations is most dangerous because of its high penetrating powerand high energy?(a) Alpha particles (b) Beta particles (c) Gamma rays (d) None of these
23. The nucleus resulting from 238U92 after successive loss of tw0 alpha and four beta particles
isa)238 TH 90 b) 230 Pu 94 c) 230 Ra 88 d) 230 U 92
24. An isotone of 76Ge32 isa) 77 Ge 32 b) 77 As 33 c) 77Se 34 d) 77Se 36
25. If Uranium (mass number 238 and atomic number 92) emits an α particle, the product hasmass number and atomic numbera) 236and 92 b) 234and90 c) 238and90 d) 236and90
26. Which of the following in an isotope of 32 Ge 76 ?
a) 33 As 77 b) 32 Ge 77 c) 34 Se 77 d) 35 Br 80
27. The atomic number of a radioactive element increases by one unit in :
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a) alpha emission d) beta emission c) gamma emission d) electron capture
28. Sulphur -35 (34.96903 a. m. u) emits a particle but no ray. The product is chlorine35(34.96885 a.m.u.). The maximum energy of the particle emitted isa) 0.00018 MeV b) 930MeV c) 0.16758 MeV d) 1.6758 MeV
29. Which of the following is a magic number?a) 54 b) 10 c) 20 d) 18
ANSWER KEY
CHAPTER # 3
PROBLEMS
1. I = 250 mA = 250 10-3 A, t = 8 sec We know that,
I =Q
t
Q = I tQ = 250 10-3 8Q = 2000 10-3 = 2 C
2. (a) Let resistance between P and Q is R PQ.3 and 3 are in series. So,
R S = 3 + 3 = 6 R S and 3 are in parallel. So,
R PQ = 33
S
S
R R
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=3 6
3 6
R PQ =18
9= 2
(b) Let resistance between A and B is R AB.
3 , R PQ and 3 resistors are in series. So,R AB = 3 + R PQ + 3R AB = 3 + 2 + 3 = 8
3. Resistance of 1 and 2 are in series. So,R S = 1 + 2 = 3
R S and 1.5 are in parallel. So,
R eq =1.5
1.5
S
S
R
R
R eq =3 1.5 4.5
3 1.5 4.5
= 1
4. When resistors are in parallel, then
R P =2 2
2 2r
R P =4
4+ r = 1 + r
Let r = Internal resistance.Current IP = 1.2 A
= IP (R P + r)
= 1.2 (1 + r)= 1.2 + 1.2 r … (i)
2
2
r
Again, when resistances are in series, thenR S = 2 + 2 = 4R eq = 4 + rIS = 0.4 A
r
2 2
+ -
Thus,
= IS (R S + r)
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= 0.4 (4 + r)= 1.6 + 0.4 r … (ii)
From equations (i) and (ii) we have, - 1.2 r = 1.2 … (i) - 0.4 r = 1.6 … (ii)
- 1.2 r = 1.2 … (i) 3 1.2 r = -4.8 … (iii)
-2 = -3.6
=3.6
1.82
volt
Again form equation (i) we get, - 1.2 r = 1.2
1.2 r = - 1.2
or, r =1.8 1.2
1.2
= 0.6 0.51.2
5. Equivalent resistance = 2 + 0.7 + 4.5 = 7.2 = 1.8 V
Current in the circuit (I) =1.8
7.2= 0.25 A
(a) Reading of ammeter = 0.25 A (b) Let terminal potential of the battery be V then,
= I (R + r) = IR + Ir
i.e., = V + Iror, V = - Ir
V = 1.8 – 0.25 2 = 1.3 volt
6. R =1
( / 4 )a and R 1 =
31
( / 2 )a
1 (31) /( / 2)
1/( / 4)
R a
R a
=3 2 3
4 2
R 1 : R = 3 : 2
7. We know that,
1 1
2 2
1
1
R
R
2 12
1
11
R
R
=16.25 75
3.25
= 375 cm
8. Net e.m.f. = 1.5 volt
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Total internal resistance,1 1 1 1 3
2 2 2 2 pr
2
3 p
r
Total resistance =7 2 9
33 3 3
Current through the resistor =1.5
0.53
V A
R
9. We have,P = 60 W, V = 250 V
P = VI
I =P
V
=60
250= 0.24 Amp
10. We have, V = 4 V, I = 0.5 A, r = 2.5 (i) Energy = V I t
E = 4 0.5 600 ( t = 10 60 = 600 sec)= 1200 Joule
(ii) As e.m.f. of battery = 4 V then,
e.m.f. = IC (R + r)
R =e.m.f
I- r
=4
2.50.5
R = 8 – 2.5 = 5.5
(iii) P = V R I= 2.75 0.5 ( V R = 5.5 0.5 = 2.75 V)= 1.375 W
Energy dissipated = 1.375
10
60= 825 Joule
11. We have,P1= 60 W, V 1 = 220 V, P2 = 60 W, V 2 = 110 V
2
1
1
1
V R
P
=220 220
60
and, 2
110 110
60
R
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1
2
220 220
60110 110
60
R
R
=
220 220 4
110 110 1
So, R 1 : R 2 = 4 : 1
12. We have,P1 = 100 W, P2 = 200 W
1 2
2 1
1, when V isconstant
R P R
R P P
1
2
200
100
R
R
So, R 1 : R 2 = 2 : 1
13. We have,R 1 = 4 , R 2 = 6 , V = 6 Volt
Req =4 6
4 6
= 2.4 (As R 1 and R 2 in parallel)
(i)2V
P =Req
=6 6
2.4
=36
2.4= 15 W
(ii) Since, potential across each resistor is same , so,2
1
1
V P
R
=6 6
94
W
and,
2
2
2
V
P R
=6 6
6
= 6 W
P1 = 9 W, P2 = 6 W
14. We have,P1 = 50 W, V = 220 V P2 = 100 W, V = 220 V If both are connected in series then their resistances add upto form an equivalentresistance. And current is same in series arrangement.
2
1
1
V R
P
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=220 220
96850
2
2
2
V R
P
220 220 484100
Req = 968 + 484 = 1452 220
0.151452
I A
Now potential drop across R 1 V 1 = I R 1 = 0.15 968 = 145.2 V
And potential drop across R 2 V 2 = I R 2 = 0.15 484 = 72.6 V
Power dissipated in R 1 = V 1I= 145.2 0.15 = 21.78 W
Power dissipated in R 2 = V 2I= 72.6 0.15 = 10.89 W
Power dissipated in first lamp is more in comparison the second because resistanceof first lamp R 1 is greater than the resistance R 2 of the second lamp.
15. 2V
P R
2V
R P
=220 220
484100
V = IR V
I R
=220
484
= 0.45 A
16. (i)1
V R
=150
12
= 75
(ii) Energy produced in 1 minute P = IV 1 60= 0.2 15 60= 180 Joule
17. Energy produced in 5 minutes P = IV 5 60= 0.5 6 5 60= 900 Joule
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18. 2
1t
V (R and P is constant)
2
1 2
2
2 1
t V
t V
2
5 200 200
220 220t
2
5 220 220
200 200t
= 6.05 minutes
19. P
I V
=24
12
2 A
Energy liberated in 20 minutes (20 60 sec)= 24 20 60= 28,800 Joule
20. kWh =1000
VIt V V IR I
R
=2
1000
V t
R
=
200 200 5
200 1000 60
= 0.0167 kWh
OBJECTIVE PROBLEMS
1. D 2. B 3. A 4. D 5. B 6. A 7. A
8. C 9. B 10. C 11. D 12. D 13. D 14. B
15. A 16. B 17. D 18. A 19. D 20. A 21. B
22. C 23. A 24. C 25. A 26. A 27. C
CHAPTER # 4
PROBLEMS
1. No. Because at any point on this line, tangent gives the direction of magnetic field. But atthe intersection point, we obtain two tangents, i.e. two directions of magnetic field, which is not
possible.
2. Yes. The face of circular coil for which the direction of current is anticlockwise behaves asNorth pole while the face having clockwise direction of current behaves as South pole.
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3. (i) By increasing the strength of current.(ii) By increasing the number of turns per unit length.(iii) By introducing a soft iron core in the solenoid.
4. When a coil is rotated in a strong magnetic field then the magnetic flux linked with coilchanges with time. This change in flux results in an induced current in the coil. Thedirection of induced current is giv en by Fleming‟s Right Hand rule.
5. When a current carrying coil is placed in a strong magnetic field then it experiences atorque on it. The direction of this torque is given by Fleming‟s Left Hand rule.
6. The speed of an electric motor depends on(i) the number of turns in the coil(ii) the strength of current(iii) the strength of the magnetic field.
7. In 1831, Michael Faraday introduced electromagnetic induction. According to him, if themagnetic flux linked with a closed circuit changes then an induced e.m.f. (induced current)is set up in that circuit. This phenomenon is called as electromagnetic induction.
8. “The induced current in a coil will appear in such a direction that it opposes the change inmagnetic flux which is responsible for its production”.
9. D.C. Motor A.C. generator(i) A D.C. motor changes electrical energy
into mechanical energy.
(ii) It works on the principle that when anelectric current is passed through aconductor placed in a magneticfield, the conductor starts moving as aresult of force acting on it.
A simple A.C. generator changes mechanical
energy into electrical energy.
It works on the principle that whenmagnetic flux passing through a closedcircuit changes, it results in the induction of an e.m.f. thereby current flows in the coil.
10. Transformers are used in electric circuit and devices which operate at a voltage other thanthe supplied voltage.No, they cannot be used with a direct current source.
11. As, NP = 800, NS = 8, EP = 220 V, ES = ?
S S
P P
N E
N E
SS P
P
NE E
N
=8
220800
= 2.2 Volt.
OBJECTIVE PROBLEMS
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1. B 2. D 3. B 4. A 5. B 6. C
7. B 8. D 9. C 10. C 11. A 12. B
13. B 14. A 15. D 16. D 17. B 18. A
19. C 20. A
CHAPTER # 5
PROBLEMS
1. (a) The total number of protons presents in the nucleus of an atom is known as the atomicnumber of that element.
(b) The total number of protons and neutrons present in the nucleus of an atom is known asthe mass number of that element.
2. The atoms of different elements which have the same mass number but different atomicnumbers, are called isobars. Eg. 11Na23 and 12Mg23.
3. Radioactivity is that nuclear phenomenon in which spontaneous emission of radiationoccurs. Such substances are called radioactive substances, e.g., Uranium, Radium,Thorium, and Polonium, etc.
4. (i) -particles are composed of helium nucleus (2He4).
-particles are composed of electrons (-1e0).
-radiations are electromagnetic radiations composed of -rays photons.
(ii) -particles have least penetrating power.
5. After or -emission, the nucleus acquires the excited state, i.e., it has excess of energy.
This excess of energy is emitted in the form of -rays photon.
6. It changes to ionised helium atom i.e. now it carries only one unit positive charge.
7. A neutron of nucleus is converted to an electron and a Proton, when a -particle is emitted.
1. 0n1 +1P1 + -1e0 + v ii. Neutron Proton Electron Antinutrino
8. (i) Atomic number = 15i. Mass number = 31
(ii) Atomic number = 16
ii. Mass number = 32
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9. 11Na24 12Mg24 + -1 0
The numbers 11 and 24 are called the atomic number and mass number respectively. They tell the number respectively. They tell the number of protons and neutrons in that element.
10. It changes to the neutral atom of helium.
11. The new nucleus formed is 86Rn
88Ra226 86Rn222 + 2He4 ( -particle)
12. Isobars have different atomic numbers, i.e., different number of protons, hence they havedifferent chemical properties.
13. P = A – 4 and Q = Z.
14. On the fission of one heavy nucleus more neutrons are produced which further disintegrateother nuclei resulting in a chain of fission is formed. Such reaction is called chain reaction.
15. because to start a fusion reaction, high temperature and pressure is required.
Objective Problems
1. A 2. B 3. D 4. C 5. A 6. C 7. B
8. A 9. D 10. C 11. C 12. D 13. B 14. A
15. A 16. D 17. A 18. A 19. C 20. A 21. C
22. B 23. C