In a radioactive decay chain- the initial nucleus is 23290Th- At the end there are 6 - - particles and 4 - - particles which are emitted- If the end nucleus- If AZX- A and Z are given by-

The correct option is D A = 208; Z = 82 ^232_90Th →^208_78Y + ^4_2He ^208_78Y →^208_82X + 4β particle ...

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Standard X

Physics

Alpha Decay

In a radioact...

Question

In a radioactive decay chain, the initial nucleus is $_{90}^{232} T h$ . At the end there are $6 α -$ particles and $4 β -$ particles which are emitted. If the end nucleus, If $_{Z}^{A} X, A$ and $Z$ are given by:

A = 208; Z = 80

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A = 202; Z = 80

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A = 200; Z = 81

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A = 208; Z = 82

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Solution

The correct option is D $A = 208; Z = 82$
$_{90}^{232} T h \to_{78}^{208} Y +_{2}^{4} H e$
$_{78}^{208} Y \to_{82}^{208} X + 4 β$ particle

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In a radioactive decay chain, the initial nucleus is 23290Th. At the end there are 6α− particles and 4β− particles which are emitted. If the end nucleus, If AZX,A and Z are given by:

The correct option is D A=208;Z=8223290Th→20878Y+42He20878Y→20882X+4β particle

In a radioactive decay chain, the initial nucleus is $_{90}^{232} T h$ . At the end there are $6 α -$ particles and $4 β -$ particles which are emitted. If the end nucleus, If $_{Z}^{A} X, A$ and $Z$ are given by:

The correct option is D $A = 208; Z = 82$
$_{90}^{232} T h \to_{78}^{208} Y +_{2}^{4} H e$
$_{78}^{208} Y \to_{82}^{208} X + 4 β$ particle