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Byju's Answer
Standard XII
Physics
Binding Energy
The binding e...
Question
The binding energy of
35
17
C
l
nucleus is 298 MeV. Find its atomic mass. Given, mass of a proton
(
m
p
)
=
1.007825
a
m
u
, mass of neutron
(
m
n
)
=
1.008665
a
m
u
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Solution
Here proton number
Z
=
17
and neutron number,
N
=
A
−
Z
=
35
−
17
=
18
Binding energy,
B
E
=
Δ
m
c
2
Mass defect ,
Δ
m
=
298
M
e
V
/
c
2
=
298
/
931.5
a
m
u
=
0.3199
a
m
u
as
(
1
a
m
u
=
931.5
M
e
V
/
c
2
)
If M be atomic mass,
Δ
m
=
(
17
m
p
+
18
m
n
)
−
M
or
M
=
(
17
×
1.007825
+
18
×
1.008665
)
−
0.3199
=
34.9597
a
m
u
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1
Similar questions
Q.
The binding energy of
17
C
l
35
nucleus is
298
MeV. Find its atomic mass. The mass of hydrogen atom
(
1
H
1
)
is
1.008143
a.m.u. Given
1
a.m.u.
=
931
MeV.
Q.
The binding energy of
17
C
I
35
nucleus is
298
M
e
V
.
Find its atomic mass. The mass of hydrogen atom
(
1
H
1
)
is
1.008143
a
m
u
and that of a neutron is
1.008986
a
m
u
.
Given
1
a
m
u
=
931
M
e
V
.
Q.
M
p
denotes the mass of a proton and
M
n
that of a neutron. A given nucleus, of binding energy
B
, contains
Z
protons and
N
neutrons. The mass
M
(
N
,
Z
)
of the nucleus is given by
(
c
is the velocity of light
)
Q.
The energy required to separate the typical middle mass nucleus
120
50
S
n
into its constituent nucleons is :
(Mass of
120
50
S
n
=
119.902199
a
m
u
; mass
of proton
=
1.007825
a
m
u
and mass of neutron
=
1.008665
a
m
u
)
Q.
Find the binding energy per nucleon for
120
50
S
n
. Mass of proton
m
p
=
1.00783
U
, mass of neutron
m
n
=
1.00867
U
and mass of tin nucleus
m
S
n
=
119.902199
U
. (take
1
U
=
931
MeV
)
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