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Byju's Answer
Standard XII
Physics
Rutherford's Model
In successive...
Question
In successive emission of α and β-particles, the number of α and β-particles should be emitted for the conversion of
92
U
238
to
82
P
b
206
are:
A
7
α
,
5
β
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B
6
α
,
4
β
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C
4
α
,
3
β
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D
8
α
,
6
β
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Solution
The correct option is
C
8
α
,
6
β
Suggest Corrections
0
Similar questions
Q.
In successive emission of
α
- and
β
- particles, the number of
α
- and
β
- particles that should be emitted for the conversion of
92
U
238
to
82
P
b
206
are
Q.
The number of
α
and
β
−
emitted during the radioactive decay chain starting from
226
88
R
a
and ending at
206
82
P
b
is
Q.
( a ) Calculate number of
α
and
β
−
particles emitted when
92
U
238
changes into radioactive
82
P
b
206
.
( b )
T
h
234
disintegrates and emits
6
β
−
and
7
α
−
particles to form a stable element. Find the atomic number and
mass number of the stable product. Also identify the element:
Q.
By the successive disintegration of
92
U
238
, the final product obtained is
82
P
b
206
, then how many number of
α
and
β
particles are emitted?
Q.
Match the following:
List-I
Series
List-II
Particles emitted
(
A
)
Thorium
(
i
)
8
α
,
5
β
(
B
)
Neptunium
(
i
i
)
8
α
,
6
β
(
C
)
Actinium
(
i
i
i
)
6
α
,
4
β
(
D
)
Uranium
(
i
v
)
7
α
,
4
β
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