$_{92}^{236} U$ atom disintegrates to $_{82}^{207} P b$ with a half-life of $10^{9}$ years. In the process it emits 7 alpha particles and n $β$ particles. Here n is

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Solution

The correct option is C 4
$23592 U \to 20782 P b + 7_{2}^{4} H e + 4_{- 1 e^{0}}$
so, $n = 4$ ( $β^{- 1}$ particles)

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23692U atom disintegrates to 20782Pb with a half-life of 109 years. In the process it emits 7 alpha particles and n β particles. Here n is

The correct option is C 423592U→20782Pb+742He+4−1e0so, n=4(β−1 particles)

$_{92}^{236} U$ atom disintegrates to $_{82}^{207} P b$ with a half-life of $10^{9}$ years. In the process it emits 7 alpha particles and n $β$ particles. Here n is

The correct option is C 4
$23592 U \to 20782 P b + 7_{2}^{4} H e + 4_{- 1 e^{0}}$
so, $n = 4$ ( $β^{- 1}$ particles)