Atoms

NTSTO
'REMEMBER
a-particlee Scattering Experiment
Geiger.Marsden's
REMEMBER
POINTSTO
suggestionof Rutherford,, iin 1911, histwo associates, H. Geiger and E. Marsden, performed
Onexperimentby bombarding a-particles (Helium nuclei Z= 2, A = 4) on a gold foil.
the
an
Observations:

oftthe a-particles pass

through the gold foil undeflected.
() Most suffered large angle deflection; some of them
small number of a-particles (1 in 8000)
(ü) Avery suffered 180° deflection.
retraced their path or
Conclusion:
hollow.
() Atom is
and
(ü) Entire positive charge
nearly whole mass of atom is
concentrated in a small centre
called nucleus of atom. Incident
beam of
(n) Coulomb's law holds good for a-particles
atomic distances. Nucleus

(tu) Negatively charged electrons Detector

are outside the nucleus.
Impact Parameter: The
perpendicular distance of initial
relocity vector of a-particle from Rutherford Scattering experiment
the nucleus, when the particle is far
away from the nucleus, is called the impact parameter. It is denoted by b. For head on approach
ol a-particle, b = 0.
Angle of from its original direction is
(6) : The angle by which a-particle is deviated
called angleScatoftering F

scattering. 1 Ze
E

4n¬, EK -cot 9
h=
b' large
wher
aphae particle.
E, is the initial kinetic energy for head on approach of A B

Impact parameter, b= 0. H
(+Ze)
N

Atoms 435
 2. Distance of Closest Approach
POINTS
TO
REME The smallest distance approach of a-paticle near heavy nucleus
of
Distance of nearest approach size of nucleus = 222
is a
measure of the
4nE, K
where E is kinetic energy of incident a-particle, Z = atomic
S. Rutheford's Atom Model number, e = electronic
Atom consists of acentral heavy nucleus
circulating around the nucdeus in circularcontaining
orbits. positive charge and
negatively
ctharge
Rutherford model could explain the neutrality of an atom,
effect: but it could not explainthe stability
(atomic spectrum).
4. Bohr's Model
Bohr modified
of an atom and the.

Rutherford atom model to explainthe line

thermiobserved
onic emission
line and chargel. ertons
spectrumphOLOel rtatiem
of an

Postulates of Bohr's Theory spectrum of

(i) Stationary Circular hydrogen.
charged nucdeus and Orbits: An atom consists of a
negatively charged central positively
electrons
nucleus in certain orbits called stationary orbits. revolve around the
The electrostatic coulomb force
provides the necessary centripetalbetween
force.
electrons and the nucleus +

14., 1 (Ze)(e) +Ze

where Z is the atomic number, m is
the mass of electrons, r =
(ii) Ouantum radius of orbit.
Condition: The hstationary orbits are those in
is an integral multiple of which angular momentum of eleron
, l.e.,
2
mUT
n = 1,2, 3,..
Integer n is called the principal
condition. quantum number. This equation is called
(iii) Transitions: The Bohr's quantum
electron does
energy (or photon) is emitted ornot radiate energy when in a stationary orbit. The
to the other. The absorbed when an electron jumps from one quantum ot
frequency of emitted or absorbed
photon is stationary orbi
hv = |E,-EA given by
This is called Bohr's
frequency condition.
Radius of Orbit and Energy of
Condition of motion of electronElectron in Orbit
in circular
orbit is
(Ze) (e)
..0
Bohr's quantum condition is
mUr n
2T
nh
Substituting this value of2Tmrvin (), we get
2Tmr 4nE,
436 Xam idea
Physics-XII
Ths
g7ves
r=
REMEMBER
POINTSTO
radiussof nth orbit by , we have
Denotng

TmZ2 ...m)
Z =].
hvdrogenatom
or
2
Tme

offirst orbit of hydrogen atomis called Bohr's radius. It is denotedi by ao

Theradius
= 0.529 X10-m = 0.529 Å
Electron
Energy ofOrbiting
mu= 1 Ze
Fromequation (i),

Kinetic energy, 4nE, 2r

I (Ze)(-e)
Potential energy, U=

1 Ze' 1 Ze
E = K+U=
Total energ) 4ne, 2r 47e
1 Ze
E = 2r

For nth orbit, writing E, for E, we have

1 Ze ...(iu)
E, =
47E, 2r,
Substituting the value of r, from (ü) in (v), we get m2²,4
...v)
E, =-
4ne, (Ehn 8en
TmZe
met ...(ui)
Rydberg constant, R=
For convenience introducing 8e~ch
1.097 x 10' m.
The value of Rydberg constant is
We have
z² Rhc ...(vü)

For hydrogen atom Z= 1,

Energy of orbiting electron in Hatom
Rhc
E, =:
13.6
E,=- -eV

Quations (iü) and (vii) indicate that radi and energies ofhydrogen like atoms (i.e., atoms containing
one electron only) are quantised.

Atoms 437
POINTSTO
RHMHu5. Energy levcls of
Hvdrogen Aom
atom(2 = ) is given (or series
Thecncrgv of cltron in hvdrogen13.6 of hydrogen The
wave
number
(i.e., reciprocal of
v= rR wavelength) of
E, =
Rhc eV;
spectrum) b relation explains successfully
the
m

when n= 1. E, = -1S.6 eV nso Continuum E> 0 The

series of lines are obtained due origin of v
to the
13.6 eV =-3.4 eV
n=7
n=6
The
innerorbit. transitic
when n =2, E, =: n=5
Continuum
13.6 n=4 n=7
when n = 3. E, = - eV =-1.5leV n=6

13.6 n=3 n=5

eV =-0.85 eV
when n = 4,
E4=-16
13.6 151el
n=4

n=3
when n = 5, E, = -eV = -0.54 eV n=2
95 Paschen se
13.6
when n=6, 36
eV = -0.38 eV n=2 series
Lyman
Balmer series
13.6
when n=7. E, = -eV =-0.28 eV
49 n=1
-13.60 ev
Fig. (a) Energy Level
Diagram
13.6
when n = , E = -eV = 0 eV n=1

If these energies are expressed by vertical lines on proper scale, Fig. (b)Series
the diagram obtained is called ihe
energy level
diagram. The energy level diagram of hydrogen atom is () Lyman series: This series is produced
separation between lines goes on decreasing rapidly with increase ofshown in fig. (a). Clearv the
n (i.e., order of orbit). The stationary orbit (i.e., n, =1). Thus for thi
series of lines of H-spectrum are shown in fig. (b).
If the total energy of electron is above zero, the
electron is free and can have any energy. Thus
there is a contúnuum of energy states above E = 0eV.
6. Hydrogen Spectrum For longest wavelength of Lyman serie
Hydrogen emission spectrum consists of 5 series.
(i) Lyman series: This lies in ultraviolet region. max
4 4
(ii) Balmer series: This lies in the visible
(iiü) Paschen series: This lies in near region. 3R 3X 1.097
infrared region.
(iv) Brackett series: This lies in mid = 1.215x10 m
(v) Pfund series: This lies in far
infrared region.
infrared region. For shortest wavelength of yman se
Hydrogen absorption spectrum consists of only Lyman series.
Explanation of Hydrogen Spectrum: n, and n,are the quantum numbers of initial and final states a Amin
E, and Ef are energies of electron in H-atom (Z =1) in
initial and final states then we haye R 1.097 X
Rhc Rhc
£, =
n? and E,= n
This is called series limit of Lyman
Energy of absorbed photon Obviously the lines of Lyman series
(ii)
Balmer series: The series is prod
Second stationary orbit (n,= 2). 1
AE = E,- E,= Rhc|
1
Ifv is the frequency of emitted V= = R
radiation, we have from Bohr's fourth postulate
V= E-E Rc Rc For Longest wavelength of Balme
h =
Rc
= R
438 Xarm idea Amax
Physics-XIl
 of wavelength) of the emitted radiation is given by
neumber
-(i.e., reciprocal
1he
=R REMEMBER
POINTS TO
explainssuccessfully the origin of various lines the spectrum of hydrogen atom.
rclation
obtained due to the transition of electron from various other orbits to a fixed
linesare
seriesof
Jhc

The
n ne
orbit.
t Continuum
0eV
n =0 -0.28 eV
n=7 -0.38 eV
n=6 -0.54 eV
n=5
Pfund series
-0.85 eV
n=4 Brackett series
-1.51 eV
n=3 Paschen series

-3.40 eV
n=2 Balmer series
s•esueu

-13.6 eV
n=1
Fig. (b) Series of H-spectrum
higher orbits to the first
This series is produced when electron jumps from
à Lyman series: Thus for this series
stationary orbit (i.e., n,=1).
where n, = 2, 3, 4, 5,...

Lymanseries n, =2
For longest wavelength of 3R
1 =
4
max

4 4 m
^max 3R
3x1.097x107
= 1215 Å
= 1.215x10 m
For shortest wavelength of Lyman series n,=0
1 =

911.6¢
x 10-7m =
min
m = 0.9l16
min 1.097x 1o?
911.6 Å
This is called series limit of ILyman series imi in= ultraviolet region. orbits to the
found jumps from higher
Obviously the lines of Lyman series are electron
i when an
Balmer series: The series is produced this series,
secondi stationary orbit (n, = 2). Thus for 5, 6,..
where n, =3, 4,

Eor =3)
Wavelength of Balmerseries (%; 5R
Longest 1 = 36 Atoms 439
^max
POINTS
TO
REMEMEER
36 36
Amay BR 5x1.097x 107" 6.563 × 10-, =
6563
For Shortest wavelength (or series limit) of Balmer series .

-
4
R 1.097 x10-7 m = 3.646 X10-7 m=3646 Å
Obviouslv the lines of Balmer series are found in the visible
region and
lines are called H, H, H,., ines respectively. first, second, third
(i) Paschen series: This series is produced when an electron jumps
from higher
stationary orbit (n, =3). orbits to the third
where n;=4, 5, 6, 7,..
For Longest wavelength of Paschen series (n, = 4)
1 7R
=

144
Nnax
144 144
7R m = 18.752 × 10'
7x 1.097X 107
For Series limit of Paschen series (n, =o)
m=18752 Å

Mmin
min R
1.097 x 10
= 8.204 x 10 m =8204 Å
Obviously lines of Paschen series are found in infrared
(iv) Brackett series: This series is region.
fourth stationary orbit (n, =4) produced when an electron jumps from higher orbitsto the
4 where n,=5, 6,7, 8, ...
(u) Pfund series:This series is
stationary orbit (n, =5) produced when an electron jumps from higher orbitsto the u

where n, = 6, 7, 8, ...
The last three series are
The series spectrum of found in infrared region.
hydrogen atom is
FOCUS
SNDILS represented in figure.
Multiple Choice Questions
Choose and write the correct
1. The
potential energyoption(s)
of an
in the
electron infollowing questions.
the second
(a) -3.4 eV excited state in hydrogen :atom
[CBSE
is
2023(55/41
(c) -1.51l eV (b) -3.02 eV
440 Xam idea (d) -6.8 eV
Physics-Xl
 Because according to Bohr's model.

E, =
13.6
and electrons having different energies belong to different evels
values of n.
So, their angular momenta will be
different, as
having dilt
nh

Q.. Define ionization energy. How would the ionizationnenergy change when
atom is replaced by a particle of mass 200times
than that of the
electron but electrohavin ning hydrthengsam
the ground state 1CBSE
the Central 2916
charge?
electron from
Ans. The minimum energy required to free the of
is known as ionization enegy.
me
i.e., E, « m
hydrogr a
8e4?"
Therefore, ionization energy will become 200 times.
Q.4. Draw the graph showing the variation of the number (N) of alphascattered
scattering angle (e) in Geiger - Marsden experiment. Infer two conclusions room parttheicles wih
CBSE 2022 (55/1/l), grTerm.
aph.
Ans. Graph: (N) 10
particles
2 10
105
scattered
10

103
of
Number
102

101

o 20 40 60 80 100 120 140 160 180

Scattering angle (in degree)
(Give ful credit if axis are marked and values are not given)
Conclusions
" Most of the alpha particles pass undeviated through the gold foil.
" Afew alpha particles, get deflected through 90 or more.
" Only about 0.14% of theincident alpha particles are reflected by large angle.
" Avery few alpha particles retrace their path.
(Any other two conclusions) [CBSE Marking Scheme 2022 (55/1/1),Trn-2]

Q.5. The ground state energy of hydrogen atom is -13.6 eV, What is the potential
[CBSE2023energy
(55/W/0)
kinetic energy of an electron in the third excited state?
Ans. For ground state, Energy (E) =-13.6 eV
For third excited state, n = 4,
E, = -13.6
-13.6
=-0.85 eV
4

448 Xam idea Physics-XI1

 KE = -E= -0.85) = 0.85 eV
and PE = -2K.E = -2 x 0.85 = - 1.7 eV.
momentum
Ilse Bohr's model of hydrogen atom to obtain the relationship betweenthe angular
and the magnetic moment of the revolving electron. (CBSE 2020 (55/5/1)]

Ans. According to Bohr's model

nh
L= Angular momentum = mur = 2
u= Magnetic moment = current Xarea of the orbit
|e|x le ur
2r 2
L mUr X 2 2m
Now, |el
le ur

2m
[CBSE Marking Scheme 2020 (55/5/1)]
obtained? Calculate the frequency
0.7. When is H, line in the emission spectrum of hydrogenatom [CBSE North 2016]
of the photon emitted during this transition.
is called H,.
Ans. The line with the longest wavelength of the Balmer series
1

where =wavelength
R= 1.097x10 m (Rydberg constant)
=3 to n = 2,
When the electron jumps from the orbit with n
We have
5
R
36

The frequency of photonemitted is given by

C 5
V==cx R
=3x108y -x 1.097 x 10 Hz
36

=4.57 x 10'4 Hz
levels in an atom. What is the frequency of
radiation
Q. 8. Adifference of 2.3 eV separates twoenergy
from the upper level to the lower level? [NCERT]
emitted when the atom makes transition
Ans. According to Bohr's postulate
E-E, = hv
radiation
. Frequency of emitted
2.3eV
y= E-E h
2.3 X1.6 x 10 = 5.55x 10 Hz
6.63 x 10-s

Atoms 449
 What names are given to the symbols b' and 0' shown here?
(b) Whhat can we say about the values ofb for (i) 0=0° (ii) 0= Iradians?
Ans. (a) Thesymbol'b represents impact parameter and'@' represents the scattering angle.

b ) 1When 8=0°,the impact parameter will be maximum and represent the atomic size.
(i) When 9=n radians, the impact parameter 'b' willbe minimum and represent the nuclear
size.
Which is easier to remove: orbital electron from an atom or a nucleon from a nucleus? [HOTS]
Ans. I is easier to remove an orbital electron from an atom. The reason is the binding energy of
orbital electron is a few electron-volts while that of nucleon in a nucleus is quite large (nearly
8MeV). This means that the removal of an orbital electron requires few electron volt energy
..hile the removal of anucleon from a nucleus requires nearly 8 MeV energy.
Wriite shortcomings of Rutherford atomic model. Explain how these were overcome by the
0.16. postulates of Bohr's atomic model. (CBSE 2020 (55/5/1)]
Ans. Two important limitations of Rutherford model are:
()According to Rutherford model, electron orbiting around the nucleus, continuously radiates
energy dueto the acceleration; hence the atom will not remain stable.
ü) As electron spirals inwards; its angular velocity and frequency change continuously, therefore
it should emit a continuous spectrum.
But an atom like hydrogen always emits a discrete line spectrum.
Bohr's postulates overcome these limitations by:
)Bohr stated that negatively charged electrons revolve around positively charged nucleus
in certain orbits called stationary orbits. The electrons does not radiate energy when in
stationary orbits.
(ü) The quantum of energy is released or absorbed when an electron jumps from one stationary
orbit to another.
0.17. Find the ratio of the longest and the shortest wavelengths amongstthe spectral lines of Balmer
series in the spectrum of hydrogen atom. [CBSE 2020 (55/4/1)]
Ans. For shortest wave length,
1

1 R
..)
A_ 4
For longest wave length,
1 -

= R ..)

Dividing equation (0) by equation (ü) we get,

5R
36

’ y:y= 9:5
5
[CBSE Marking Scheme 2020 (55/4/1)]

Atoms 453
 Nuclei

POINT'STO
REMEMBER
1. Composition of
Nucleus REMEMBER
and
POINTS
almost entire
TO
mass.
entire positive charge
The atom consists of central nucleus, containing
According to accepted model the nucleus is composed of protons and neutrons. The proton was
on nitrogen in accordance with the
discovered by Rutherford by bombardment of a-particles
following equation:
He + }H
Proton
(a-particle) Oxygen
(Nitrogen)
mass number and subscripts (in the base)
denote the atomic
superscripts (on the top) denote the the
The or ,x, where Ais the mass number and Z is
number.Symbolically a nucide is written as X
atomic number.
beryllium in
neutron was discovered by |. Chadwick by the bombardment of a-particles on
The
accordance with
+
'Be + He (Carbon) (Neutron)
(Beryllium) (a-particle)
is 1.
charge) particle and its mass number
Aneutron is neutral (zero while the number of nucleons
nucleus is called atomic number (Z)
The number of protons in a mass number>atomic number
neutrons) is called the mass number (A). In general
(Le., protons +
where A = Z).
(except for hydrogen nucleus
used for artificial disintegration.
Since neutron is neutral, it is
atom of mass number A is
the radius of the nucleus of an
2. Size of Nucleus
experimental observations,
According to
where Ro = 1.2x10" m=1.2 fm
R= RA3
3. Atomic Masses terms of atomic mass unit represented by amu
atoms, nuclei, etc., are expressed in
The masses of taken as standard.
of C-12 atom is
or u'. or this mass
mass ofC- 12 atom
|u 12
ko
= l.660565x 102
proton (m,) = 1.007276 u
mass of
= 1.008665 u
mass of neutron (m,)
mass of electron (m) =0.000549 u
and Isotones
4. Isotopes, Isobars called isotopes.
The nucides having the same atomic number (Z) but different mass number (A) are
The nuclides having the same mass number (A), but different atomic number (Z) are called isobars.
The nucides having the same number of neutrons (4-Z) are called isotones.

Nuclei 465
 3. Mass Energy Equivalence Relation
According to Einstecin, the mass and energy are equivalent i.., mass can
be
and vie-versa. The mass cnecrgy cquivalence relation is E = mc.
Accordingl, Ikg mass is cquivalent to energy = I x (3 × 10°) = 9x 10 irul.
converted into enevg
1
and Iamu kg mass is equivalent to energy 931 MeV.
6.02 x 1026
6. Mass Defect
It is observed that the mass of a nucleus is always less than the
mass of
protons t neutrons). This difference of mass is called the mass constituent
nudeus, m, = the mass of proton and m, = mass of neutron, then defect.
the massLetdefert
(Z, A) be
the nucleons le
Am = Mass of nucleons - Mass of nucleus nao
= Zm,+(4 - Z)m, M,nucleus
7. Binding Energy per Nucleon
This mass defect is in the form of binding energy of nucleus, which is
nudeons into a smallnucleus. responsible for
.. Binding energy of nucleus = (Am) c
binding the
and (Am)e
Binding energy per nucleon A
8. Nature of Nuclear Forces
The protons and neutrons inside the nucleus are held
attractive forces cannot be gravitational since forces on together by strong attractive forces. The
repulsion between protons > > alrati
gravitational force between protons. These forces are
forces. The nuclear forces are strongest in nature, shortshort
range
range attractive forces called nuclr
the force between proton-proton is the same as the and charge independent, therefore
force between neutron-neutron or proton:
neutron.
Yukawa tried to explain the existence of these forces,
not have independent existence between nucleus. The accordingly the proton and neutron do
proton and neutron are interconverible
through negative and positive T-mesons, i.e.,
Proton Neutron and Neutron Neutron
The existence of meson gives rise to meson field which gives rise to attractive
nuclear forces.
The mass of n-meson = 273 X mass of electron.
9. Nuclear Reaction

When a beam of monoenergetic particles (e.g., a-rays, neutrons etc.)collides with a stable nuclets,
the original nucleus is converted into anucleus of new element. This process is called a nuclea
reaction. Atypical nuclear reaction is
a + X Y+b
where a iS incident energetic particle, X is target nucleus. Yis residual nucleus and bis ougos
particle. This reaction in compact formn is expressed as
X(a, b) Y
In a nuclear reaction mass number, electric charge, linear momentum, angular momentum and

total energy are always conserved. The energy of reaction is

10. Nuclear Fission

Q= (M, + M) c- (M, + My¢
Ihe splitting of heavy nucleus into two or more fragments of comparable masses,are
witnbombarded
ded on o
release of energy is called nuclear fission. For when slow neutrons ar
gU9, the fission takes place according to example,
reaction
8 U + Ba + 2Kr + 3n) + 200 MeV
36
(slow neutron)
466 Xam idea Physics-X|
 lo nuclear fission the sumn of massesbefore reaction is greater than the sum ofmasses after reaction,
he difference inmass being released in the lorm of fission energ.
Remarks: REMEMBER
ONTS TO
Llmay be pointed out that it is not necessary that in cach fission of uranium, the two fragments
Ba and Kr areforned but they may be any stable isotopes of middle weight atoms. The
most probable divisicon is into two fragments containing about 40% and 60% of the original
Ducleus with the emission of 2or 3 neutrons per fission.
) The fission of U² takes place by fast neutrons.
I1. Nuclear Fusion
release
The phenomenon of combination of twoor more light nuclei to form a heavy nucleus with greater
of cnormous amount of energy iscalled nuclear fusion. The sum of masses before fusion is
than the sunm of masses after fusion, the difference in mass appearing as fusion energy.
For example, the fusion of two deuterium nuclei into helium is expressed as
H +H He + 21.6 MeV.
where
Thus, fusion process occurs at an extremely high temperature and high pressure as in sun
temperature is 10' K.
Remarks:
within a distance of
1. For the fusion to take place, the component nuclei must be brought between
10" m. For this they must be imparted high energies to overcome the repulsive force
nuclei. This is possible when temperature is enormously high.
2. The principle of hydrogen bomb is also based in
nuclear fusion.
possible cycles:
3. The sourceof energy of sun and other star is nuclear fusion. There are two
(a) Proton-proton cycle:
H+ H 'H + B + v. (Neutrino) + Energy
}H + HHe + n+ Energy
He + He}He + H + H + Energy
H+H + H+H He + 2B + 2v + Energy (26.7 MeV)
Net resultis
(b) Carbon-nitrogen cycle:
H+C N+ Energy
N C + B + v (neutrino)
C+ H - N + (Energy)
N+ H IO + Energy
o N + ,B" + v (neutrino)
N + }H C + He + Energy
Net resultis H + H+ H +H He t 2p + 2 + Energy (26.7 MeV)
The proton-proton cycle occurs at a relatively lower temperature as compared to carbon
nitrogen cycle which has a greater elliciency at higlher temperature.
AL the sun whose interior temperature is about 2 x 10° K, the proton-proton cycle has more
chances for occurrence.

Nuclei 467
 Vo (ume athydrgen ot om 3.313
OCupied y its wcles 4g423
B395 xo-16
4240

[Topper's Answer 20221

Q. 6. () Distinguish between isotopes and isobars.
(ii) Two nuclei have different mass numbers A, and A,. Are these nuclei
necessarily theeisobars
of the same element? Explain. [CBSE 2022 (55/3/), (55/3/3), Term-21
Ans.

TSOToPES ISn BARS.

+The atoms mhich - Theatoms hich
have the Same atamidhavediterent
Ihombes but diffeoent ato mic nombegs but
mass nuMbes ae ISotg ame mo0Ssnumbeo
Yatomg of
JSob aS diteesen ekrete
ISotofes haé some ISobcabs have
Dumbeo of poatons in diffeoent oumb ex of
them pootons io themn
not conStant is Consant
2

(iTsotopes have Same atonic numbexbut diffezent

MASS Qmhes TA othes sdsiotapes haue. equa

hiven -AucleiA andc t have diceTeO maes.

4umbeas* Theseyhuo Auclei can he isotape.O0ly it.
they h0Uefhe Same otomic Aumbes.
he and havedifEeaent_mass Dumbea
buthe4 aaeAot 1sotapes
H andHhave dieeent nceS nbmb
CADe iSotop es
ns:hns kwonucleihavedì ffezett Mass Qùmbess
44adAaEhey angat ne ce ssaiy ne 1S0 to
QktheSamelement |1opper'sAnswer2022)

476 Xam idea Physics-XI|

 OR
Draw a diagram to show the variation of binding energy per nucleon with mass number
for different nuclei and mention its two features. Why do lighter nuclei usually undergo
nuclear fusion? [CBSE 2023 (55/2/1))

Ans.
The variation of binding energy per nucleon versus mass number is shown in figure.
Inferences from graph
relatively small binding
1. The nuclei having mass number below 20 and above 180 have
energy and hence they are unstable.
maximum binding energy - 88 MeV
2. The nudei having mass number 56 and about 56 have
and so thev are most stable.

have peaks, e.g., He,C,"0; this indicates that these nuclei are relatively
3. Some nuclei
more stable than their neighbours.
Explanation of constancyofbinding energy: Nudear force is short ranged, so every nucdeon
() nucleon remains constant.
binding energy per
interacts with its neighbours only, therefore two
of nuclear fission: When a heavy nucleus (4 > 235 say) breaks into
(ii) Explanation nucleons
binding energy per nucleon increases i.e,
lighter nuclei (nuclear fission), the fission.
tightly bound. This implies that energy would be released in nucdear
get more
9.0
o56
Fe'
UR38
MeV) 8.0c'
He Ni4
(in 7.0
Nucleon
6.0

5.0
per
4.0
Energy
3.0
Binding
2.0
H2
1.0

200 220 240

100 120 140 160 180
00 20 40 60 80
Mass Number

When two verylight nuclei(4 s 10) join to form a heavy

fusion: than the
(iii) Explanation of nuclear energy per nucleon of fused heavier nucleus more
nucleus, the binding is released in
lighter nuclei, so again energy would be
of
binding energy per nucleon
nuclear fusion.

terms:
Q.2. (i) Define the following number, (c) Mass number, (d) Nuclear
mass
Atomic
(a) Nucleons, (b) isotones?
What are (a) isotopes, (b) isobars and (c) considered as nucleons.
(ii)
nucleus i.e.,protons and neutrons are
The constituents of the atomic
Ans. () (a) Nucleons: protons present in the nucleus is called
The number of
(b) Atomicnumber: element. It is denoted by
Z.
number of the