Question 160: Atoms & Nuclei

Question

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 :

Accepted Answer

In the given hypothetical fission reaction:
$$^{236}_{92} X ightarrow \, ^{141}_{56} Y + \, ^{92}_{36} Z + 3 R$$
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:
$$236 = 141 + 92 + 3 	imes A_R$$
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

Answer

Proton

Answer

Neutron

Answer

Electron

Answer

$$\gamma$$-radiations

Atoms & Nuclei

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 :

Choose the correct answer: