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Beta Decay

How many α ...

Question

How many $α$ and $β$ -particles will be released in the nuclear reaction given below?

$_{90} {T h}^{232} ⟶_{82} {P b}^{208}$

A

6 α

and

4 β

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B

6 α

and

6 β

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C

4 α

and

6 β

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D

4 α

and

4 β

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Solution

The correct option is A $6 α$ and $4 β$
$_{90} T h^{232} \to_{82} P b^{208} + Y (_{2} α^{4}) + X (_{- 1} β^{0})$

Mass number:
$232 = 4 Y + 208 + 0 X$
$4 Y = 24$
$Y = 6$

Atomic number:
$90 = 82 + 2 Y - X$
$90 = 82 + 12 - X$
$X = 4$

Hence, $α -$ particles $= Y = 6$

$β -$ particles $= X = 4$

Hence, option A is correct.

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Similar questions

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α

β^{-}

emitted during the radioactive decay chain starting from

\begin{matrix} 22688 \end{matrix} R a

\begin{matrix} 20682 \end{matrix} P b

Q. In the radioactive disintegration series

^{232} T h_{90} \to^{208} P b_{82}

α a n d β

decay, the total number of

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particles emitted are:

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	$(i v)$ $7 α, 4 β$

{_{90} T h}^{232} ⟶ {_{82} P b}^{208}

. The number of

α

β

- particles emitted during the above reaction is:

α

β

are the roots of the equation

2 x^{2} - 7 x + 8 = 0

, then the equation whose roots are

(3 α - 4 β

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