{ }_{92}^{235} U atom disintegrates to { }_{8 | vedclass

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Question

(c)

Given, $7 \alpha$ particles are emitted from ${ }_{92}^{235} U$.

$\Rightarrow { }_{92}^{235} U \longrightarrow{ }_{78}^{207} A +7\left({ }_2

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(c)

Given, $7 \alpha$ particles are emitted from ${ }_{92}^{235} U$.

$\Rightarrow { }_{92}^{235} U \longrightarrow{ }_{78}^{207} A +7\left({ }_2^4 He \right)$

Now, $n \beta^{-}$particles are emitted to produce ${ }_{82}^{207} Pb$

$\Rightarrow { }_{78}^{207} A \longrightarrow{ }_{82}^{207} Pb +n\left({ }_{-1}^0 \beta\right)$

Conservation of atomic number, gives

$78=82-n \Rightarrow n=82-78=4$

So, $4 \beta^{-}$particles are emitted in given nuclear reaction.

${ }_{92}^{235} U$ 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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