"In a hypothetical fission reaction ${ }_{92} X^{236} \rightarrow{ }_{56} \mathrm{Y}^{141}+{ }_{36} Z^{92}+3 R$ The identity of emitted particles (R) is :"

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In the given hypothetical fission reaction:

$^{236}_{92} X \rightarrow \, ^{141}_{56} Y + \, ^{92}_{36} Z + 3 R$

We need to determine the identity of particles denoted by $ R $. Let's use the conservation of charge and mass number (nucleon number) to identify $ R $.

First, for the conservation of nucleon number (mass number), we have:

$236 = 141 + 92 + 3 \times A_R$

Where $ A_R $ is the mass number of $ R $. This simplifies to:

$236 = 233 + 3A_R$

$3A_R = 236 - 233$

$3A_R = 3$

$A_R = 1$

Next, we use the conservation of charge (atomic number), we have:

$92 = 56 + 36 + 3Z_R$

Where $ Z_R $ is the atomic number of $ R $. This simplifies to:

$92 = 92 + 3Z_R$

$3Z_R = 92 - 92$

$3Z_R = 0$

$Z_R = 0$

Since the particle $ R $ has a mass number of 1 and atomic number of 0, it must be a neutron.

So, the identity of the emitted particles $ R $ is:

Option B: Neutron

"

In a hypothetical fission reaction

${ }_{92} X^{236} \rightarrow{ }_{56} \mathrm{Y}^{141}+{ }_{36} Z^{92}+3 R$

The identity of emitted particles (R) is :

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In a hypothetical fission reaction ${ }_{92} X^{236} \rightarrow{ }_{56} \mathrm{Y}^{141}+{ }_{36} Z^{92}+3 R$ The identity of emitted particles (R) is :

Solution

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More Atoms and Nuclei questions

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In a hypothetical fission reaction

${ }_{92} X^{236} \rightarrow{ }_{56} \mathrm{Y}^{141}+{ }_{36} Z^{92}+3 R$

The identity of emitted particles (R) is :