QJEEM24Ph64f5: JEE Main 2024 Physics - Class 12 JEEM24Ph64f5 Solution: In a hypothetical fission reaction 92 X 236 → 56 Y 141 + 36 Z…

In the given hypothetical fission reaction:

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

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