nuclei
01
NUCLEAR RADIUS
1/3
1/3
R∝ A
R = R0 A
P+n
NUCLEAR COMPOSITION
Na23 mass number = p+n=A
02
Atomic number = Z
Z = proton = 11
NUCLEAR VOLUME
mass of nucleus< (Z)mp +
V∝A
Neutron = 23-11=12
27
If radius of the 13
Al nucleus is taken to
125
Te nucleus is nearly
be RAl, then the radius of 53
(a)
53
RAl
13
(b)
5
RAl
3
(b)
3
13
RAl (a)
RAl
5
53
1/3
NUCLEAR FORCE
Y+
Nuclear force is
short ranged
r>0.8 fm - attractive
r<0.8 fm - repulsive
Nuclear force is
charge independent
F= F = F
P-P
n-n
P-n
As Ebn increases the nucleus
becomes more stable
mR > mP
B.E / nucleon given
-
2
0 = vy (A-4) - v 4
vy =
-
v 4
A-4
m x 931 MeV
P2 ;
K.E
2m
n
p+
A
A
Z
Z+1
X
Q = K.E + K.Ey
1
m
4
xQ
A
-
Y+
-
Atomic number increases
by one and mass number
remains same
K.E of particle is more
than daughter nucleus
A-4
K.E =
xQ
A
p
n+
A
A
Z
Z-1
+
Y+
A= A0e
+
Atomic number decreases
by one and mass number
remains same
heavy nuclei (A>170)
Shortcut
t1/2 =
- λt
N
(b) The helium nucleus has more kinetic (d) The helium nucleus has more momentum
than the thorium nucleus.
energy than the thorium nucleus.
0.693
Decayed
Undecayed
N0
2
N0
2
t1/2
3N0
4
N
N0
= 02
4
2
2t1/2
7N0
8
N
N0
= 30
8
2
3t1/2
(a)
Time
No. of undecayed nuclei:
λ
N
0
N = 2t/t1/2
Two radioactive materials A and B have decay
constants 10 λ and λ , respectively. If initially
they have the same number of nuclei, then the
ratio of the number of nuclei of A to that of B
will be 1/e after a time:
(c) The helium nucleus has less momentum
than the thorium nucleus.
N0
2n
t = nt1/2
Time at which ratio of nuclei will
be 1/e
N
t = 2.303 log 0
Ionizing power : Alpha > Beta > Gamma
(d) 6, 8
(d) 32, 12
Ebn is lower for both light nuclei (A<30) and
A=-dN
dt
A= λN
A nucleus of uranium decays at rest into
nuclei of thorium and helium. Then:
(d) 16, 6
For nuclei of middle mass number (30<A<170)
Ebn is a constant
ED
Activity
In the uranium radioactive series the initial nucleus
is 92U238 and final nucleus is 82Pb206. When the uranium
nucleus decays to lead, the number of a-particles
emitted is and the number of -particles emitted =
(d) 8, 6
EC
Undecayed =
λ
(a) The helium nucleus has less kinetic
energy than the thorium nucleus.
C + D
Nt= N0e-λt (N No. of undecayed nuclei)
No change in atomic number & mass number
Penetrating power : Gamma > Beta > Alpha
EB
The curve has a maximum of about
8.75 MeV for A=56
A4
dN =λN
dt
Decay
Decay
X
EA
A3
Q value = (ECA3 + EDA4) - (EAA1 + EBA2)
Gamma Decay
A-A‛
n =
4
No.of alpha decays
Decay
+
A2
Decay law at radio activity
+
Decay
A1
A + B
B.E / nucleon
Beta Decay
particle + Q
Momentum Conservation
K.Ey =
mP
B.E =
Alpha Decay
A-4
Z-2
Q= [mx-my-m ]C
K.E =
C + D
Ebn = Eb /A
Q valve = B.Ep - B.ER
K.E
Strongest force
existing in nature
M (in amu) x 931 MeV
m = mR - mp
Independent of A
It is same for all atoms
ρ=2.3 x 1017 kg/m3
X
B.E =
mR
(A-Z)mn
NUCLEAR DENSITY
A
Z
Mc
A + B
M=[Zmp +(A-Z)m n]-mnucleus
03
1/3
nucleus
B.E =
BINDING ENERGY CURVE
{
{
11
BINDING ENERGY
PER NUCLEON
BINDING ENERGY
MASS DEFECT
R0 =1.2 fm
1
1
(b)
9λ
11λ
Half-lives of two radioactive elements A and B are 20
minutes and 40 minutes, respectively. Initially, the
samples have equal number of nuclei. After 80 minutes,
the ratio of decayed number of A and B nuclei will be:
(a) 1 : 4
(b) 5 : 4 (c) 1 : 16
(d) 4 : 1
11
1
(c)
(d)
10 λ
10λ
NUCLEAR FISSION
t2
N1
N2
Age of rock
Time interval between 33% (1/3) &
67% (2/3) is t1/2
N
t1 - t2 = 2.303 log N1
λ
2
Time interval between 20 % and 80% decay , or b/w 40 % and 85 % decay (t2 - t1) is 2 t1/2
The half-life of a radioactive substance is 30 min.
The time (in minutes) taken between 40% decay & 85%
decay of the same radioactive substance is:
(a) 15
(b) 60
(c) 45
(d) 30
X
Y (Y is stable)
Method 1
Nx m
=
Ny n
N0=Nx+Ny=m+n
m+n
t= 2.303 log ( m
λ
(
Undecayed
t1
Method 2
N0 a m+n
=2 = m
N
Two deuterons undergo nuclear fusion to form a Helium nucleus.
Energy released in this process is: (given binding energy per nucleon
for deuteron-1.1 MeV and for helium=7.0 MeV)
Age= at1/2
(a) 30.2 MeV (b) 32.4 MeV (c) 23.6 MeV (d) 25.8 MeV
The half life of a radioactive isotope 'X' is 20 years.
It decays to another element "Y" which is stable.
The two elements 'X' and 'Y' were found to be in
the ratio 1:7 in a sample of a given rock. The age
of the rock is estimated to be
(a) 40 years (b) 60 years
(c) 80 years (d) 100 years
A radioactive nucleus (initial mass number A and atomic number Z
emits 3 α - particles and 2 positrons. The ratio of number of
neutrons to that of protons in the final nucleus will be
(a)
A-Z-8
Z-4
(b)
A-Z-4
Z-8
(c)
A - Z - 12
Z-4
(d)
A-Z-8
Z-2
1
0
n+
235
92
NUCLEAR REACTOR
Multiplication Factor k=1
89
1
→ 144
U → 236
92 U
56 Ba+ 36 Kr+3 0 n
NUCLEAR FUSION
critical
4 H + 2e
+ 6 + 26.7 MeV
2 He + 2
Moderator : water,heavy water (D2 O),
Four hydrogen atoms combine to form an 24 He
graphite and beryllium oxide.
atom with the release of 26.7 MeV of energy
Control rods : Boron, cadmium
Achieved at very high temperature in order to
Coolant : CO2, water, nitrogen
overcome electrostatic repulsion
1
1
-
4
CHAPTER
13
Characteristics of Nucleus
The nucleus of an atom is its center. Most of the mass of an
atom is concentrated in the nucleus.
™ One atomic mass unit is defined as (1/12)th of the mass of
the 126C isotope. It is represented by the symbol u and it is the
average mass of a nucleon.
™ The stability of any nucleus depends on the number of protons
and neutrons. For small nuclei to be stable, the number of
protons must be roughly equal to the number of neutrons. As
the number of protons increases, however, more neutrons are
needed to maintain stability.
™ Nuclei of different elements have different sizes because the
mass number (A) for different elements is different. Density
of nucleus is the almost same for all elements.
™ Average radius of nucleus may be written as,
Nuclei
™
™
R = R0A1/3, R0 = 1.1 × 10–15 m
Mass Defect
The sum of the individual masses of the separated protons and
neutrons exceeds the mass of the stable nucleus by an amount
∆M. This difference in mass is known as the mass defect of the
nucleus.
Mass-energy Relation
According to mass-energy relation, the mass of a body is a measure
of its energy content.
Equivalence of mass and energy, E = mc2
Note: 1 u = 1.66 × 10–27 kg ≡ 931.5 MeV
Binding Energy
The binding energy of a nucleus is the energy that would
have to be provided to split a nucleus into its individual
nucleons. Binding energy of nucleus Z X A of mass M, is
given by BE = (Zmp + Nmn – M)c2 and the mass defect,
∆m = Zmp + Nmn – M.
™ The average energy required to remove a nucleon from the
nucleus to infinite distance is known as the binding energy
per nucleon.
™ Nuclei with high binding energies per nucleon are very stable
as it takes a lot of energy to split them. Nuclei with lower
binding energies per nucleon are easier to split.
™
In order to become more stable, unstable nuclei tend to release
some of their energy. Releasing this energy would decrease
the amount of energy they contained, and therefore increase
the amount of energy that must be added to them to split them
apart.
Q Value
™
Energy released in any nuclear reaction or collision is called
Q value of the reaction.
™
For nuclear reaction, A + B → C + Q (Energy)
The energy of reaction Q is given by,
Q = (mA + mB – mC)c2 = BEC – BEA – BEB
If Q is positive, energy is released and products are more
stable in comparison to reactants.
If Q is negative, energy is absorbed and products are less
stable in comparison to reactants.
™
Nuclear Force
The strong nuclear force is the force that holds nucleons
together in the nucleus of an atom. It acts only over very short
distances.
™ Inside a nucleus, the nucleons are very close. The pull of
the strong nuclear force is much greater than the push of
the protons repelling each other, and therefore the nucleus
remains intact.
™ Adjacent neutrons experience no electrostatic repulsion
between each other. There is only the strong force of
attraction) between them.
™ Two protons do repel each other when they are brought
together, but in the nucleus they are so close to each other
that the force of repulsion is overcame by the even stronger
nuclear force.
™
Nuclear Fission
™
By bombarding a particle on a heavy nucleus (A > 230), it
splits into two or more light nuclei. In this process certain
mass disappears which is obtained in the form of energy
(enormous amount)
A+p→B+C+Q
The energy released in a fission reaction comes from the
difference between the mass of the original nucleus and the
combined mass of the fission fragments.
™ A chain reaction occurs when neutron, emitted from the decay
of one atom, initiate fission in surrounding nuclei.
™ An uncontrolled chain reaction occurs when every free
neutron goes on to produce another fission reaction. It
occurs in nuclear bombs and releases an enormous amount
of energy.
™ A controlled chain reaction occurs when only some of the
emitted neutrons are able to induce further fission reactions,
and the remaining neutrons are absorbed by a material that is
not fissionable. In this situation energy can be released at a
constant rate.
™ Enrichment is the process of increasing the percentage
of 235U in a sample of uranium. Enrichment is important
because naturally occurring uranium does not have a high
enough percentage of 235U to sustain a chain reaction.
™
In β-decay the resulting daughter nucleus has the same
number of nucleons as the parent, but has one less neutron
and one more proton.
™ β-particles are very light when compared to α particles. They
travel at a large range of speeds-from that of an α- particle up
to 99% of the speed of light.
™
™
A X → A Y + b– + v–
Z
™
A nuclear phenomenon in which an unstable nucleus undergoes
a decay is known as radioactivity. A decay equation is a
representation of a decay reaction. It shows the changes occurring
in nuclei undergoing decay and lists the products of the decay.
The nucleus remaining after an atom undergoes radioactive decay
is called a daughter nucleus. The daughter nucleus is more stable
than the original nucleus. Generally, there are three types of
radioactive decays,
(i) α decay
(ii) β– and β+ decay
(iii) γ decay
a decay
An α-particle is a relatively slow-moving decay product
consisting of two protons and two neutrons. It is the nucleus
of helium and so can be written as 42He. α-particles carry
positive charge.
A–4
4
™ α-decay process: A
Z X → Z–2Y + 2He
A–4
4
2
™ Q-value is, Q = [m(A
ZX) – m( Z–2Y) – m( 2He)]c
™ For the above shown decay kinetic energy of α-particle will
m
be given by, K α = Y Q
mX
™
β– and β+ decay
™
30
A β-particle is a fast-moving electron that is ejected from an
unstable nucleus. The electron is produced when a neutron
transforms into a proton and an electron.
β+ decay
A X → AY + b+ + v
Z
Z–1
AY) – m ]c2
Q – value = [m(AZX) – m( Z–1
e
AY) are atomic masses
where m(AZX), m( Z–1
™
Electron capture (K-capture): When atomic electron is
captured by nucleus, X-rays are emitted.
AX + 0 e → AY + v
Z
–1
Z–1
AY)]c2
Q – value = [m(AZX) – m( Z–1
™
Radioactivity
Z+1
A Y)]c2
Q – value = [m(AZX) – m( Z+1
Nuclear Fusion
It is the phenomenon of fusing two or more light nuclei to
form a single heavy nucleus.
A + B → C + Q (Energy)
™ The product (C) is more stable than reactants (A and B) and
mC < (mA + mB).
™ Mass defect, ∆m = [(mA + mB) – mC] amu
™ Energy released is E = (∆m) × 931 MeV
β–decay
AY) are atomic masses.
where m(AZX) – m( Z–1
g decay
A γ-ray is the packet of electromagnetic energy released when
a nucleus is formed in excited state after α or β-decay which
releases energy to come down to ground state. γ-rays travel at the
speed of light, carry no mass or charge, and are not deflected by
electric or magnetic fields.
Law of Radioactive Decay
The rate of disintegration is directly proportional to the number of
radioactive atoms present at that time i.e., rate of decay ∝ number
of nuclei.
Rate of decay = λ (number of active nuclei)
i.e.,
dN
= –λN.
dt
where λ is called the decay constant.
If N0 is the number of parent nuclei at t = 0. The number that
survives at time t is N = N0e–λt
N
N0
N e−λt
N0
0
Probability of a nucleus for survival of time t=
=
e – λt
=
Half-life
™
™
A half-life is the time taken for half of a group of unstable
nuclei to decay. In other words, it is the time during which the
number of undecayed atoms in the sample becomes half the
total number of atoms present initially in the sample. Half-life
is represented by the symbol T1/2.
ln 2 0.693
Half life: T=
=
1/ 2
λ
λ
Number of nuclei present after n half lives i.e. after a time t
= nT1/2, N = N 0
2n
JEE (XII) Module-4 PW
™
A radioactive nucleus can decay by two different processes
having half lives t1 and t2 respectively. Effective half-life of
1 1 1
+ .
nucleus will be given by =
t t1 t2
Mean or Average Life
The ratio of the total life time of all the atoms of the element
to the total number of atoms present initially in the sample of
the element is known as the mean or average life. Mean life is
represented by the symbol τ. It is equal to the reciprocal of the
decay constant (λ) of the element.
P Nuclei
W
Average life: τ=
T1/2
1
=
λ 0.693
Activity
Activity of sample:
R = λN = R0e–λt
™ Unit: 1 Bq = 1 decay/s, 1 curie = 3.7 × 1010 Bq, 1 rutherford
= 106 Bq
™ Activity per unit mass is called specific activity.
A0
™ Activity after n half lives:
2n
™
31
Given that mass of proton = 1.00813 amu, mass of neutron is 1.00894 amu and mass of -particle is 4.00388
amu, the binding energy of alpha particle is
(A)
28.172 MeV
(B)
27.172 MeV
(C)
13.52 MeV
(D)
56.321 MeV
(A)
BE of -particle = ቀ𝐴 − 𝑍)𝑚𝑛 + 𝑍𝑚𝑝 − 𝑀 𝐻𝑒 × 931.5
= (2 × 1.00894 + 2 × 1.00813 – 4.00388) 931.54
= 0.03026 amu × 931.5 meV/amu
BE = 28.18 MeV
भौकाल of the Topic
Topic: Binding energy
Similar Questions asked years :- 2017
Given that mass of proton = 1.00813 amu, mass of neutron is 1.00894 amu and mass of -particle is 4.00388
amu, the binding energy of alpha particle is
(A)
28.172 MeV
(B)
27.172 MeV
(C)
13.52 MeV
(D)
56.321 MeV
(A)
BE of -particle = ቀ𝐴 − 𝑍)𝑚𝑛 + 𝑍𝑚𝑝 − 𝑀 𝐻𝑒 × 931.5
= (2 × 1.00894 + 2 × 1.00813 – 4.00388) 931.54
= 0.03026 amu × 931.5 meV/amu
BE = 28.18 MeV
भौकाल of the Topic
Topic: Binding energy
Similar Questions asked years :- 2017
The separation energy of neutrons is defined to be the energy required to remove a neutron from a nucleus.
Obtain the neutron separation energies of the nuclei 20Ca41 and 13Al27 from the following data:
М1 = 1.008665 u, М(20Са40) = 39.962591 u,
M(20Ca41) = 40.962278 u, M(13Al26) = 25.986895 u and M(13A127) = 26.981541 u
(A)
8.363 MeV & 13.06 MeV respectively
(B)
13.6 MeV & 8.363 MeV respectively
(C)
Both have equal separation energy for neutron
(D)
None of these
(A)
Neutron separation energy Sn of a nucleus ZXA is
Similarly,
given by
 Sn (13Al27) = [M(13Al26) + mn – М(13Al27)] u
Sn = [M(ZXA–1) + mn – M(ZXA)]c2
= [25.986895 + 1.008665 – 26.981541] u
= [{M(ZXA–1) + Zme} + mn – {M(ZXA) + Zme}]c2
= 0.014019 u
= [M(ZXA–1) + mn – M(ZXA)]c2
= 0.014019 × 931 MeV = 13.06 MeV
 Sn(20Ca41) = [M(20Ca40) + mn – M(20Ca41)]c2
Sn = 13.06 MeV for Al
= [39.962591 + 1.008665 – 40.962278] u
= (40.981256 – 40.962278) u
= 0.0089781 = 0.008978 × 931 MeV
भौकाल of the Topic
Sn = 8.363 MeV for Ca
Topic: Nuclei
Similar Questions asked years :- 2017
Find the Binding energy per neucleon of 120
50Sn. Mass of proton mp = 1.00783 u, mass of neutron
mn = 1.00867 u and mass of tin nucleus mSn = 119.902199 u.
(Take 1u = 931 MeV)
(A)
7.5 MeV
(B)
9.0 MeV
(C)
8.0 MeV
(D)
8.5 MeV
(D)
m = (50mp + 70mm) – (msn)
= (50 × 1.00783 + 70 × 1.00867) – (119.902199)
= 1.096
Binding energy = (m)c2 =(m) × 931 = 1020.5631
Binding energy 1020.5631
=
= 8.5 MeV
Nucleon
120
भौकाल of the Topic
Topic: Mass-Energy Equivalence
Similar Questions asked years :- 2020