Phy 310 chapter 9

Dr Ahmad Taufek Abdul Rahman
School of Physics & Material Studies, Faculty of Applied Sciences, Universiti Teknologi MARA Malaysia, Campus of Negeri Sembilan
Studies,
Sciences,

is d fi d
i defined as a
physical process in
which there is a
change in identity
of an atomic
nucleus.
nucleus

CHAPTER 9: Nuclear reaction
(2 Hours)
Four types of
nuclear reaction:

1

DR.ATAR @ UiTM.NS

PHY310 NUCLEAR REACTION

Learning Outcome:
L
i
O t
9.1 Nuclear reaction (1 hour)
At the end of this chapter, students should be able to:
State the conservation of charge (Z) and nucleon
number (A) in a nuclear reaction.
Write and complete the equation of nuclear reaction.
reaction
Calculate the energy liberated in the process of nuclear
reaction

2

DR.ATAR @ UiTM.NS


9.1 Nuclear reaction
9.1.1 Conservation of nuclear reaction
Any nuclear reaction must obeyed conservation laws stated
below:
Conservation of relativistic energy (kinetic and rest
energies):

∑

∑

relativistic energy =
gy
before reaction

relativistic energy
gy
after reaction

Conservation of linear momentum:

∑

∑

linear momentum =
before reaction

linear momentum
after reaction

Conservation of angular momentum:

∑

∑

angular momentum
=
before
b f reaction
ti

angular momentum
after reaction
ft
ti
3

DR.ATAR @ UiTM.NS


Conservation of charge (atomic number Z):

∑

atomic number Z
before reaction

=∑

atomic number Z
after reaction

Conservation of mass number A:

∑

mass number A
before reaction

=∑

mass number A
after reaction

However, it is very hard to obey all the conservation laws.
Note:
N t
The most important of conservation laws should be obeyed
by every nuclear reaction are conservation of charge (atomic
atomic
number )and of mass number
number.

4

DR.ATAR @ UiTM.NS


9.1.2 Reaction energy (Q)
Energy is released (liberated) in a nuclear reaction in the form of
kinetic energy of the particle emitted the kinetic energy of
energ
emitted,
energ
the daughter nucleus and the energy of the gamma-ray
gammaphoton that may accompany the reaction.
The energy is called the reaction OR disintegration energy (Q).
It may be calculated by finding the mass defect of the reaction
where

Mass defect

=∑

mass of nucleus
before reaction

∆m = mi − mf

−∑ products after reaction
mass of nucleus
(9.1)
(9.1)

The reaction energy Q is the energy equivalent to the mass
defect ∆m of the reaction, thus

Q = (∆m )c

2

(9.2
(9 2)
(9.2)
2)
Speed of light in vacuum
5

DR.ATAR @ UiTM.NS


Note:
If the value of ∆m OR Q is positive the reaction is called
positive,
exothermic (exoergic) in which the energy released in the
form the kinetic energy of the product.
If the value of ∆m OR Q is negative, the reaction is called
g
endothermic (endoergic) in which the energy need to be
absorbed for the reaction occurred.

9.1.3 Radioactivity decay
is defined as the phenomenon in which an unstable nucleus
disintegrates to acquire a more stable nucleus without
absorb an external energy
energy.
The disintegration is spontaneous and most commonly
involves the emission of an alpha particle (α OR 4 He ), a beta
0
0 2
particle (β OR −1 e ) and gamma-ray (γ OR γ ). It also
0
release an energy Q k
l
known as di i t
disintegration energy
ti energy.
6

DR.ATAR @ UiTM.NS


Example 1 :
Polonium nucleus decays by alpha emission to lead nucleus can be
represented by the equation below:
212
Po→208 Pb+ 4 He + Q
84
82
2

Calculate
a. the energy Q released in MeV.
b. the wavelength of the g
g
gamma-ray p
y produced.
(Given mass of Po-212, mPo=211.98885 u ; mass of Pb-208,
mPb=207.97664 u and mass of α particle , mα=4.0026 u)
212
Solution :
α decay
Po→208 Pb+ 4 He + Q
84
82
2
before
decay

∑Z = ∑Z
i

f

after
decay

and

∑A =∑A
i

f

7

DR.ATAR @ UiTM.NS


Solution :
a. The mass defect (difference) of the reaction is given by

∆m = mi − mf
= mPo − (mPb + mα )
= 211.98885 − (207.97664 + 4.0026)
∆m = 9.61× 10−3 u

The energy released in the decay reaction can be calculated by
gy
y
y
using two method:
2
st method:
1 u = 1.66 ×10−27 kg
1
Q = ∆m c

( )

(

in kg

)(

)

∆m = 9.61 × 10 −3 1.66 × 10 −27
−29
= 1.5953 × 10 kg
− 29
8
Q = 1.5953 × 10
3.00 ×10
Q = 1.436 × 10 −12 J

(

)(

)

2

8

DR.ATAR @ UiTM.NS


Solution :
a. Thus the energy released in MeV is

1.436 × 10 −12
Q=
−13
1.60 × 10
Q = 8.98 MeV

2nd method:

Q = (∆m )c

2

1 MeV = 1.60 ×10−13 J

1 u = 931.5 MeV/c 2

in u

⎡ ⎛ 931.5 MeV/ c 2 ⎞⎤ 2
= ⎢∆m⎜
⎥c
⎜
1u
⎠⎦
⎣ ⎝
⎡
⎛ 931 .5 MeV/c 2 ⎞⎤ 2
−3
⎟⎥ c
= ⎢ 9.61 × 10 u ⎜
⎜
⎟
1u
⎢
⎝
⎠⎥
⎣
⎦
Q = 8.95 MeV

(

)

9

DR.ATAR @ UiTM.NS


Solution :
b. The reaction energy Q is released in form of gamma-ray where
its wavelength can be calculated b appl ing the Planck’s
a elength
calc lated by applying
hc
quantum theory:

E=

=Q
λ
hc
λ=
Q

(6.63 ×10 )(3.00 ×10 )
=
−34

Note:

1.436 × 10
−13
λ = 1.39 × 10 m

8

−12

The radioactive decay only occurred when the value of ∆m
OR Q is positive
positive.
10

DR.ATAR @ UiTM.NS


Example 2 :

(

)

A nickel-66 nucleus 66 Ni decays to a new nucleus by emitting a
28
beta particle.
particle
a. Write an equation to represent the nuclear reaction.
p (a)
b. If the new nucleus found in part ( ) has the atomic mass of
65.9284 u and the atomic mass for nickel-66 is 65.9291 u, what
is the maximum kinetic energy of the emitted electron?
(Given mass of electron, me =5.49 × 10−4 u and c =3.00 × 108 m s−1)
Solution :
a. N l
Nuclear reaction equation must obey th conservation of atomic
ti
ti
t b the
ti
f t i
number and the conservation of mass number.
66
66
0
28 Ni→ 29 X + −1 e + Q

β decay

11

DR.ATAR @ UiTM.NS


Solution :
b. Given m Ni = 65 .9291 u ; mX = 65 .9284 u
The mass defect (difference) of the reaction is given by

∆m = mi − mf
= mNi − (mX + me )
N
= 65.9291 − 65.9284 + 5.49 × 10−4
−4
∆m = 1.51× 10 u

(

)

If the reaction energy is completely convert into the kinetic
energy of emitted electron, therefore the maximum kinetic energy
of the emitted electron is given by

K max = Q
2
= (∆m )c

(

K max

−4

)(

= 1.51 × 10 1.66 × 10
= 2.26 × 10 −14 J

− 27

)(3.00 ×10 )

8 2

12

DR.ATAR @ UiTM.NS


Example 3 :
Table 9.1 shows the value of masses for several nuclides.
Nuclide
4
2 He
23
11 Na
27
13 Al

Mass (u)
4.0026
22.9898
22 9898

26.9815
Table 9.1
Discuss whether i i possible f
Di
h h it is
ibl for 27 Al to emit spontaneously an
i
l
13
alpha particle.
Solution :
If
by

27
13 Al

emits an alpha particle, the α decay would be represented
27
13 Al

26.9815 u

→

23
11 Na

22.9898 u

+

4
2 He

4.0026 u

26.9924 u
Since the total mass after the reaction is greater than that before
13
the reaction, therefore the reaction does not occur.

DR.ATAR @ UiTM.NS


9.1.4 Bombardment with energetic particles
is defined as an induced nuclear reaction that does not
occur spontaneously;
occ r spontaneo sl it is ca sed b a collision bet een a
caused by
between
nucleus and an energetic particles such as proton, neutron,
alpha particle or photon
photon.
Consider a bombardment reaction in which a target nucleus X is
bombarded by a particle x, resulting in a daughter nucleus Y,
an emitted particle y and reaction energy Q:

X + x → y + Y+ Q

sometimes this reaction is written in the more compact form:
target (parent)
nucleus

X(x, y )Y

daughter nucleus

bombarding emitted
particle
p
particle
The calculation of reaction energy Q has been discussed in
section 9.1.2.
14

DR.ATAR @ UiTM.NS


Examples of bombardment reaction:

(α , p )17 O
8
7
Li+1 H →2 4 H + Q OR 7 Li(p,α )4 He
3
1
2 He
3
2
10
1
B+ 0 n → 7 Li+ 4 He + Q OR 10 B(n,α )7 Li
5
3
2
5
3

14
4
17
1
7 N + 2 He→ 8 O+1 H + Q

Example 4 :

OR 14 N
7

17

A nitrogen nucleus 14 N is converted into an oxygen nucleus 8 O
7
and a proton if it is bombarded by an alpha particle carrying certain
amount of energy.
a. Write down an expression for this nuclear reaction, showing the
atomic number and the mass number of each nuclide and
particle emitted.
b.
b Calculate the minimum energy of the alpha particle for this
reaction to take place.
(Given mp =0.16735×10−26 kg; mα =0.66466 ×10−26 kg ; mass of
0.16735 10
0.66466 10
nitrogen nucleus , mN =2.32530×10−26 kg; mass of oxygen nucleus,
15
mO =2.82282×10−26 kg ; c =3.00×108 m s−1)

DR.ATAR @ UiTM.NS


Solution :
a. The expression represents the nuclear reaction is
14
7

+4 H → 17 O + 1 H + Q
N 2 He
8
1

b. The mass defect of the reaction is

∆m = mi − mf
= (mN + mHe ) − (mO + mH )

(

−26

−26

)

= 2.32530 ×10 + 0.66466 ×10
− (2.82282 ×10−26 + 0.16735 ×10−26 )

∆m = −2.1×10

−30

kg
k

Therefore the minimum energy of the alpha particle for this
reaction to take place is

K min = Q

K min = (∆m )c
8
−30
= 2.1×10
3.00 ×10
−13
K min = 1.89 ×10 J
2

(

)(

)

2
16

DR.ATAR @ UiTM.NS


Exercise 9 1 :
E
i 9.1
Given c =3.00×108 m s−1, mn=1.00867 u, mp=1.00782 u,
1.
1

Complete the following radioactive decay equations :
a. 8 Be→4 He +
4

2

[

]

b. 240 Po→97 Sr +139 Ba +
94

38

56

( )+ [

c. 236 U→131 I + 3 1 n
92

53

d. 11 Na →−1 e +
29

0

[

e. 47 Sc∗ →47 Sc +
21

21

f. 19
f 40 K →40 C +
20 Ca

[

0

[

[

]
]

]

]

]

17

DR.ATAR @ UiTM.NS


Exercise 9 1 :
E
i 9.1
2.

Calculate the energy released in the alpha decay below:
238
92

U→ Th + He + Q
234
90

4
2

(Given mass of U 238 mU=238.050786 u ; mass of Th 234
(Gi
f U-238,
238 050786
f Th-234,

mTh=234.043583 u and mass of α particle , mα=4.002603 u)

ANS. : 6.87×10−13 J
ANS 6 87×
6.87
87×
3. The following nuclear reaction is obtained :
14
7

N + n → C+ H + 0.55 MeV
1
0

14
6

1
1
14
6

Determine the mass of C in atomic mass unit (u).
(Given the mass of nitrogen nucleus is 14.003074 u)
(Gi
th
f it
l
i 14 003074 )
ANS. : 14.003872 u

18

DR.ATAR @ UiTM.NS


Learning Outcome:
L
i
O t
9.2 Nuclear fission and fusion (1 hour)
At the end of this chapter, students should be able to:
Distinguish th processes of nuclear fi i and fusion.
Di ti
i h the
f
l
fission d f i
Explain the occurrence of fission and fusion in the form
of graph of binding energy per nucleon.
g p
g
gy p
Explain chain reaction in nuclear fission of a nuclear
reactor.
Describe th process of nuclear fusion in the sun.
D
ib the
f
l
f i i th

19

DR.ATAR @ UiTM.NS


9.2 Nuclear fission and fusion
9.2.1
9 2 1 Nuclear fission
is defined as a nuclear reaction in which a heavy nucleus
splits into two lighter nuclei that are almost equal in mass
p
g
q
with the emission of neutrons and energy
energy.
Nuclear fission releases an amount of energy that is greater
than the energy released in chemical reaction
reaction.
Energy is released because the average binding energy per
nucleon of the fission products is greater than that of the
parent.
parent
It can be divided into two types:
spontaneous fission – very rarely occur.
occur
induced fission – bombarding a heavy nucleus with slow
neutrons or thermal neutrons of low energy (about 10−2 eV).
This
Thi type of fission i the i
f fi i is h important process i the energy
in h
production.
20

DR.ATAR @ UiTM.NS


For
F example, consider th b b d
l
id the bombardment of 235 U b slow
t f 92 by l
neutrons. One of the possible reaction is
235
1
U + 0 n →236 U∗ →85 Br +148 La
92
92
35
57

1
+30 n + Q

Nucleus in the excited state.
The reaction can also be represented by the diagram in Figure
9.1.
91
85
35 Br
1
0n

1
0n

1
0n
235
92 U

1
0n

236 ∗
92 U

Figure 9.1
91
Other possible reactions are:

235
1
U + 0 n →236 U∗ →89 Kr +144 Ba
92
92
36
56

148
57 La
1
+30 n + Q

235
1
1
U + 0 n →236 U∗ →94 Sr +139 Xe +30 n
92
92
38
54

+Q

21

DR.ATAR @ UiTM.NS


Most f th fi i f
M t of the fission fragments (daughter nuclei) of the uraniumt (d
ht
l i) f th
i
235 have mass numbers from 90 to 100 and from 135 to 145 as
shown in Figure 9.2.

Figure 9.2

22

DR.ATAR @ UiTM.NS


Example 5 :
Calculate the energy released in MeV when 20 kg of uranium-235
undergoes fission according to
235
1
U + 0 n →89 Kr +144 Ba
92
36
56

1
+30 n + Q

(Given the mass of U-235 =235.04393 u, mass of neutron
=1.00867 u, mass of Kr-89 =88.91756 u, mass of Ba-144
=143.92273 u and NA =6.02×1023 mol−1)
143.92273
6.02 10
Solution :
The mass defect (difference) of fission reaction for one nucleus U235 is

∆m = mi − mf
= (mU + mn ) − (mKr + mBa + 3mn )
= (235.04393 + 1.00867) −
(88.91756 + 143.92273 + 3 ×1.00867)
∆m = 0.1863 u
23

DR.ATAR @ UiTM.NS


Solution :
The energy released corresponds to the mass defect of one U-235
is
s
Q = ∆m c 2

(

)

⎡
⎛ 931 .5 MeV/c 2 ⎞⎤ 2
⎟⎥ c
= ⎢(0.1863 u )⎜
⎜
⎟
1u
⎢
⎝
⎠⎥
⎣
⎦
Q = 174 MeV
235 × 10−3 kg of uranium-235 contains of 6.02 × 1023 nuclei
20
⎞
20 kg of urainum-235 contains of ⎛
6.02 × 10 23

⎜
−3 ⎟
⎝ 235 × 10 ⎠

(

)

= 5.12 × 10 25 nuclei

Therefore
Energy released
by 20 kg U-235

(

)

= 5.12 × 10 25 (174 )
27

= 8.91 × 10 MeV
24

DR.ATAR @ UiTM.NS


Example 6 :
A uranium-235 nucleus undergoes fission reaction by bombarding it
with a slow neutron The reaction produces a strontium 90 nucleus
neutron.
strontium-90
90
Sr , a nucleus X and three fast neutrons.

( )
38

a. Write down the expression represents the fission reaction.
p
p
b. If the energy released is 210 MeV, calculate the atomic mass of
nucleus X.
(Given the mass of U 235 =235.04393 u, mass of neutron
(Gi
h
f U-235 235 04393
f
=1.00867 u and mass of Sr-90 =89.90775 u)
Solution :
a. The expression represents the fission reaction is
235
1
U + 0 n →90 Sr +143X
92
38
54

1
+ 30 n + Q

25

DR.ATAR @ UiTM.NS


Solution :
The energy released of 210 MeV equivalent to the mass defect for
U 35 s
U-235 is
2

Q = (∆m )c
⎡
⎛ 931 .5 MeV/c 2 ⎞⎤ 2
⎟⎥ c
210 = ⎢(∆m )⎜
⎜
⎟
1u
⎢
⎝
⎠⎥
⎣
⎦

∆m = 0.22544 u

Therefore the atomic mass of the nucleus X is given by

∆m = mi − mf
∆m = (mU + mn ) − (mSr + mX + 3mn )
0.22544 = (235.04393 + 1.00867) −
(89.90775 + mX + 3 × 1.00867)
mX = 142 .8934 u
26

DR.ATAR @ UiTM.NS


9.2.2
9 2 2 Chain reaction
is defined as a nuclear reaction that is self- sustaining as a
selfresult of the products of one fission reaction initiating a
subsequent fission reaction.
b
t fi i
reaction
ti
Figure 9.3 shows a schematic diagram of the chain reaction.

Stimulation 9.1

Figure 9.3

27

DR.ATAR @ UiTM.NS


From Figure 9.3, one neutron initially causes one fission of a
uranium-235 nucleus, the two or three neutrons released can go
on to cause additional fissions, so the process multiples.
p
p
This reaction obviously occurred in nuclear reactor.
Conditions to achieve chain reaction in a nuclear reactor :
Slow neutrons are better at causing fission – so uranium
are mixed with a material that does not absorb neutrons but
slows them down.
The fissile material must has a critical size which is defined
as the minimum mass of fissile material that will sustain
a nuclear chain reaction where the number of neutrons
produced in fission reactions should balance the
number of neutron escape from the reactor core
core.
The
Th uncontrolled chain reactions are used i nuclear weapons –
ll d h i
i
d in
l
atomic bomb (Figure 9.4).
The controlled chain reactions take place in nuclear reactors
p
(Figure 9.5) and release energy at a steady rate.
28

DR.ATAR @ UiTM.NS

Figure 9.4
94


Figure 9.5
95

29

DR.ATAR @ UiTM.NS


9.2.3 Nuclear fusion
is defined as a type of nuclear reaction in which two light
nuclei f se
n clei fuse to form a heavier n cle s with the release of
hea ier nucleus ith
large amounts of energy
energy.
The energy released in this reaction is called thermonuclear
gy
energy.
Examples of fusion reaction releases the energy are
2
3
1
H + 2 H→ 2 He+ 0 n + Q
1
1
2
H + 2 H→3 H +1 H + Q
1
1
1
1

The two reacting nuclei in f
fusion reaction above themselves
have to be brought into collision.
As bo nuclei a e pos e y c a ged there is a s o g
s both uc e are positively charged e e s strong
repulsive force between them, which can only be overcome
if the reacting nuclei have very high kinetic energies.
These high kinetic energies imply temperatures of the order
of 108 K.
30

DR.ATAR @ UiTM.NS


At these elevated temperatures however fusion reactions
temperatures,
are self sustaining and the reactants are in form of a plasma
(i.e. nuclei and free electron) with the nuclei possessing
sufficient energy to overcome electrostatic repulsion forces
forces.
The nuclear fusion reaction can occur in fusion bomb and in the
core of a star.
Deuterium-tritium f i i other example of f i reaction
D
i
ii
fusion is h
l f fusion
i
where it can be represented by the diagram in Figure 9.6.
Deuterium
Tritium
2
1H

Figure 9.6

Alpha p
p particle
4
2 He

3
1H

Fusion
reaction

Neutron
N t
1
0n

2
1
H+ 3 H→4 He+ 0 n
1
1
2

+Q

Stimulation 9.2
31

DR.ATAR @ UiTM.NS


Example 7 :
A fusion reaction is represented by the equation below:
2
H + 2 H→3 H +1 H
1
1
1
1

Calculate
a. the energy in MeV released from this fusion reaction,
b. the energy released from fusion of 1.0 kg deuterium,
(Given mass of proton =1.007825 u, mass of tritium =3.016049 u
(Gi
f
1 007825
f ii
3 016049
and mass of deuterium =2.014102 u)
Solution :
a. The mass defect of the fusion reaction for 2 deuterium nuclei is

∆m = mi − mf
= (mD + mD ) − (mT + mp )
= (2.014102 + 2.014102) − (3.016049 + 1.007825)
−3
∆m = 4.33 × 10 u
32

DR.ATAR @ UiTM.NS


Solution :
a. Therefore the energy released in MeV is

Q = (∆m )c 2
⎡
⎛ 931 .5 MeV/c 2 ⎞⎤ 2
−3
⎟⎥ c
= ⎢ 4.33 × 10 u ⎜
⎜
⎟
1u
⎢
⎝
⎠⎥
⎣
⎦
Q = 4.03 MeV

(

)

b. The mass of 2 deuterium nuclei is 4 ×10−3 kg.
4 × 10−3 kg of deuterium contains of 6.02 × 1023 nuclei
1.0 kg of deuterium contains of ⎛ 1.0 ⎞ 6.02 × 10 23

⎜
−3 ⎟
⎝ 4 × 10 ⎠

(

)

= 1.505 × 10 26 nuclei

Therefore
Energy released from
=
1.0 kg deuterium

(1.505 ×10 )(4.03)
26
6

26

= 6.07 × 10 MeV

33

DR.ATAR @ UiTM.NS


9.2.4 Nuclear fusion in the sun
The sun is a small star which generates energy on its own by
means of nuclear f i in it i t i
f
l
fusion i its interior.
The fuel of fusion reaction comes from the protons available in
the sun.
The protons undergo a set of fusion reactions, producing
isotopes of hydrogen and also isotopes of helium. However, the
helium nuclei th
h li
l i themselves undergo nuclear reactions which
l
d
l
ti
hi h
produce protons again. This means that the protons go through
a cycle which is then repeated. Because of this proton-proton
cycle, nuclear fusion in the sun can be self sustaining.
The set of fusion reactions in the proton-proton cycle can be
illustrated by Figure 9 7
9.7.

34

DR.ATAR @ UiTM.NS


positron (beta plus)
1
1
2
0
1 H +1 H→ 1 H + 1 e + v + Q
2
3
H +1 H→ 2 He +
1
1

neutrino

γ +Q

3
3
4
1
1
2 He+ 2 He→ 2 He+1 H +1 H

+Q

Figure 9.7
Fi
97
The amount of energy released per cycle is about 25 MeV.
Nuclear f i occurs i the i
N l
fusion
in h interior of the sun b
i
f h
because the
h
temperature of the sun is very high (approximately 1.5 × 107 K).

35

DR.ATAR @ UiTM.NS


9.2.5 Comparison between fission and fusion
Table 9.2 shows the differences between fission and fusion
reaction.
reaction
Fission

Fusion

Splitting a heavy nucleus into two C bi
Combines t
two small nuclei t f
ll
l i to form
small nuclei.
a larger nucleus.
It occurs at very high temperature
It occurs at temperature can be
(108 K).
controlled.
Difficult to controlled and a
Easier to controlled and
sustained controlled reaction has
sustained.
sustained
not yet been achieved.
Table 9.2
The similarity between the fission and fusion reactions is both
reactions produces energy
energy.
Graph of binding energy per nucleon against the mass number
in Figure 9.8 is used to explain the occurrence of fission and
36
fusion reactions.

DR.ATAR @ UiTM.NS


Binding energy per
n
nucleon (MeV/nu
ucleon)

Greatest stability

Fission
The falling part of the binding energy curve
shows that very heavy elements such as
uranium can produce energy by fission of their
nuclei to nuclei of smaller mass number
number.
Fusion
The rising part of the binding energy curve
shows that elements with low mass
number can produce energy by fusion
fusion.

Figure 9.8
Mass number A

37

DR.ATAR @ UiTM.NS


Exercise 9 2 :
E
i 9.2
Given c =3.00×108 m s−1, mn=1.00867 u, mp=1.00782 u,
1.
1

Complete the following nuclear reaction equations:
3
]→2 He+ 4 He
3
2
b. 58 Ni+ 2 H → [
]+1H
28
1
1

a. 6 Li +

[

1
1
c. 235 U + 0 n →138 Xe +50 n +
92
54

[

]

(

)
_____(n, p )16 N
7

d. 9 Be α , ____ 12 C
4
6
e.
2.

Calculate the energy released in joule for the following fusion
1
reaction: 2 H + 2 H→ 4 He+ 0 n
1
1
2
(Given the mass of deuterium =3.345×10−27 kg, mass of
tritium =5 008×10−27 kg mass of He = 6 647×10−27 kg and
=5.008×10 27 kg,
6.647×10 27
mass of neutron =1.675×10−27 kg)
38
ANS. : 2.8×10−12 J
2.8×

DR.ATAR @ UiTM.NS


All The Best
ll h
t
Final Exam – Oct 2012

39

Phy 310 chapter 9

Recomendados

Recomendados

Mais conteúdo relacionado

Mais procurados

Mais procurados (20)

Semelhante a Phy 310 chapter 9

Semelhante a Phy 310 chapter 9 (12)

Mais de Miza Kamaruzzaman

Mais de Miza Kamaruzzaman (20)

Último

Último (20)

Phy 310 chapter 9