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Byju's Answer
Standard XII
Chemistry
Work Done in Isothermal Reversible Process
Complete the ...
Question
Complete the following nuclear equations :
14
7
N
+
4
2
Hc
→
17
8
O
+
.
.
.
.
.
.
9
4
Be
+
4
2
He
→
12
6
C
+
.
.
.
.
.
.
9
4
Be
(
p
,
α
)
.
.
.
.
.
.
.
.
30
15
P
→
30
14
Si
+
.
.
.
.
.
.
3
1
H
→
3
2
He
+
.
.
.
.
.
.
43
20
Ca
(
α
,
.
.
.
.
)
→
46
21
Sc
A
( a )
1
1
H
,
( b )
1
0
n
,
( c )
6
3
Li
,
(
d
)
0
+
1
e
,
( e )
0
−
1
e
( f ) p (proton)
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B
( a )
1
1
H
,
( b )
1
0
n
,
( c )
6
32
Li
,
(
d
)
0
−
1
e
,
( e )
0
−
1
e
( f ) p (proton)
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C
( a )
1
1
H
,
( b )
1
0
n
,
( c )
6
−
3
Li
,
(
d
)
0
−
1
e
,
( e )
0
−
1
e
( f ) p (proton)
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D
( a )
1
1
H
,
( b )
1
0
n
,
( c )
6
3
Li
,
(
d
)
0
−
1
e
,
( e )
0
+
1
e
( f ) p (proton)
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Solution
The correct option is
A
( a )
1
1
H
,
( b )
1
0
n
,
( c )
6
3
Li
,
(
d
)
0
+
1
e
,
( e )
0
−
1
e
( f ) p (proton)
The balanced nuclear equations are as shown:
14
7
N
+
4
2
Hc
→
17
8
O
+
1
1
H
9
4
Be
+
4
2
He
→
12
6
C
+
1
0
n
9
4
Be
(
p
,
α
)
6
3
Li
30
15
P
→
30
14
Si
+
0
+
1
e
3
1
H
→
3
2
He
+
0
−
1
e
43
20
Ca
(
α
,
.
.
.
.
)
→
46
21
Sc
+
p
(
p
r
o
t
o
n
)
In these equations, the sum of the atomic numbers on LHS is equal to the sum of the atomic numbers on RHS.
Similarly,
the sum of the mass numbers on LHS is equal to the sum of the mass numbers on RHS.
Suggest Corrections
0
Similar questions
Q.
Complete the following nuclear equations.
(a)
14
7
N
+
4
2
He
→
17
8
O
+
.
.
.
.
.
(b)
9
4
Be
+
4
2
He
→
12
6
C
+
.
.
.
.
.
(c)
9
4
Be(p,\alpha)
.
.
.
(d)
30
15
P
→
30
14
S
+
.
.
.
.
.
(e)
3
1
H
→
3
2
He
+
.
.
.
.
.
(f)
43
20
Ca(\alpha ,.....)
→
46
21
Sc
Q.
Which table correctly matches the numbered equations below with the correct nuclear process?
1)
1
1
H
+
2
1
H
→
3
2
H
e
+
γ
2)
241
95
A
m
→
237
93
N
p
+
4
2
H
e
3)
235
92
U
+
1
0
n
→
139
56
B
a
+
94
36
K
r
+
3
1
0
n
4)
138
53
I
→
138
54
X
e
+
0
−
1
e
+
¯
¯
¯
v
Q.
If
I
1
=
∫
1
0
e
−
x
cos
2
x
d
x
,
I
2
=
∫
1
0
e
−
x
2
cos
2
x
d
x
I
3
=
∫
1
0
e
−
x
2
d
x
,
I
4
=
∫
1
0
e
−
x
2
d
x
then
Q.
14
7
N
+
1
0
n
→
14
6
C
+
1
1
H
is written as
Q.
The missing term in
1
+
1
p
+
0
−
1
e
⟶
1
0
n
+
?
, is:
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