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Standard IX
Chemistry
Question
In the nuclear decay given below;
A
Z
X
→
A
Z
+
1
Y
→
A
−
4
Z
−
1
B
∗
→
A
−
4
Z
−
1
B
The particles emitted in the sequence are:
γ
,
β
,
α
β
,
γ
,
α
β
,
α
,
γ
α
,
β
,
γ
A
α
,
β
,
γ
B
β
,
α
,
γ
C
β
,
γ
,
α
D
γ
,
β
,
α
Open in App
Solution
Verified by Toppr
⋅
α
−
decay
Z
X
A
α
→
Z
−
2
X
A
−
4
⋅
β
−
decay
Z
X
A
β
→
Z
+
1
X
A
⋅
γ
−
decay consists of electromagnetic radioations.
Now, the given decay series.
Z
X
A
β
→
Z
+
1
Y
A
α
→
Z
−
1
B
A
−
4
γ
→
Z
−
1
B
A
−
4
∴
The correct sequence.
β
,
α
,
γ
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4
Similar Questions
Q1
In the nuclear decay given below;
A
Z
X
→
A
Z
+
1
Y
→
A
−
4
Z
−
1
B
∗
→
A
−
4
Z
−
1
B
The particles emitted in the sequence are:
View Solution
Q2
If
α
,
β
,
γ
are the roots of
a
x
3
+
b
x
2
+
c
x
+
d
=
0
and
∣
∣ ∣ ∣
∣
α
β
γ
β
γ
α
γ
α
β
∣
∣ ∣ ∣
∣
=
0
,
α
≠
β
≠
γ
, then find the equation whose roots are
α
+
β
−
γ
,
γ
+
α
−
β
,
β
+
γ
−
α
View Solution
Q3
lf
α
,
β
,
γ
are the roots of
x
3
+
2
x
−
3
=
0
, then the transformed equation having the roots
α
β
+
β
α
,
β
γ
+
γ
β
,
γ
α
+
α
γ
is obtained by taking
x
=
View Solution
Q4
α
,
β
,
γ
are roots of
x
3
−
2
x
2
−
x
+
2
=
0
. Centroid of triangle with vertices
(
α
,
β
,
γ
)
,
(
β
,
γ
,
α
)
,
(
γ
,
α
,
β
)
is
View Solution
Q5
cos
(
α
+
β
+
γ
)
+
cos
(
α
−
β
−
γ
)
+
cos
(
β
−
γ
−
α
)
+
cos
(
γ
−
α
−
β
)
=
View Solution