KCET · Physics · Nuclear Physics
The end product of decay of \( { }_{90} T h^{232} \) is \( { }_{82} P b^{208} . \) The number of \( \alpha \) and \( \beta \) particles emitted are
respectively
- A \( 6,0 \)
- B \( 3,3 \)
- C \( 4,6 \)
- D \( 6,4 \)
Answer & Solution
Correct Answer
(D) \( 6,4 \)
Step-by-step Solution
Detailed explanation
(D)
\( { }_{90} T h^{232} \rightarrow{ }_{82} P b^{208} \)
Mass number changes by \( 24 \) and hence \( 6 \) a particles are emitted (as \( 1 \) a particle emitted decreased mass number by \( 4 \) )
then proton number should decrease by \( 12 \), but change in proton number is by \( 8 \) and hence \( 4 \beta \) particleshould be
emitted so that proton number increased by \( 4 \)
\( { }_{90} T h^{232} \rightarrow{ }_{82} P b^{208} \)
Mass number changes by \( 24 \) and hence \( 6 \) a particles are emitted (as \( 1 \) a particle emitted decreased mass number by \( 4 \) )
then proton number should decrease by \( 12 \), but change in proton number is by \( 8 \) and hence \( 4 \beta \) particleshould be
emitted so that proton number increased by \( 4 \)
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