The { }_{92}^{235} U atom disintegrates to { | vedclass

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(C) The initial nucleus is ${ }_{92}^{235} U$ and the final nucleus is ${ }_{82}^{207} Pb$.The emission of $7 \ \alpha$ particles results in a change

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$4$

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(C) The initial nucleus is ${ }_{92}^{235} U$ and the final nucleus is ${ }_{82}^{207} Pb$.The emission of $7 \ \alpha$ particles results in a change in mass number and atomic number as follows:Mass number change: $7 	imes 4 = 28$.Atomic number change: $7 	imes 2 = 14$.Let the intermediate nucleus be ${ }_{Z}^{A} X$. After emitting $7 \ \alpha$ particles:$A = 235 - 28 = 207$$Z = 92 - 14 = 78$Now,$n \ \beta^{-}$ particles are emitted to reach the final state ${ }_{82}^{207} Pb$:${ }_{78}^{207} X \longrightarrow { }_{82}^{207} Pb + n({ }_{-1}^{0} \beta)$By the conservation of atomic number:$78 = 82 - n$$n = 82 - 78 = 4$Therefore,$4 \ \beta^{-}$ particles are emitted.

The ${ }_{92}^{235} U$ atom disintegrates to ${ }_{82}^{207} Pb$ with a half-life of $10^9 \ yr$. In the process,it emits $7 \ \alpha$ particles and $n \ \beta^{-}$ particles. Here,$n$ is

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