Answer

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**Hint:**We know that for a reaction to take place, the mass number and atomic number on the product side must be equal to that on the reactant side. So, they compared the equations for mass number and atomic number on both sides, the value of alpha and beta can be calculated.

**Complete step by step solution**

The given reaction is:

$ _{92}{{\text{U}}^{238}}{{\to }_{82}}\text{P}{{\text{b}}^{214}}+\text{a }{{\text{ }}_{2}}\text{H}{{\text{e}}^{4}}+\text{b }{{\text{ }}_{-1}}{{\text{e}}^{0}} $

We assume that “a” number of (alpha) $ \alpha $ particles and “b” number of $ \beta $ (beta) particles are emitted.

Now, we know that for every reaction, the mass number and atomic number must be conserved, that is, mass number and atomic number on the product side must be equal to that on the reactant side.

Firstly, on equating the mass numbers on both sides, we get

$ \begin{align}

& 238=214+4\text{ a}+0 \\

& 238=214+4\text{ a} \\

& \text{ 4 a}=24 \\

& \text{ a}=\dfrac{24}{4} \\

& \text{ a}=6 \\

\end{align} $

Therefore, 6 alpha particles are emitted

Now, on equating the atomic numbers on both sides, we get:

$ \begin{align}

& 92=82+2\text{ a}-\text{b} \\

& \text{92}-\text{82=2 a}-\text{b} \\

& \text{92}-82=2\times 6-\text{b} \\

\end{align} $

$ \begin{align}

& \text{92}-82-12=-\text{b} \\

& -\text{b}=-2 \\

& \text{b=2} \\

\end{align} $

Therefore, 2 beta particles are emitted.

**So, the correct option is (A).**

**Note**

Alpha particles consist of 2 protons and 2 neutrons bound together into a particle identical to $ \text{H}{{\text{e}}^{4}} $ nucleus.

They are generally produced in the proven of alpha decay, The electric charge is $ +2\text{ e} $ , mass is $ 6\cdot 6446\times {{10}^{-27}}\text{ kg} $ and spin=0.

Beta particles are a high energy, high speed electron or positron which is emitted by the radioactive decays of an atomic nucleus during the process of beta decay. It has charge $ -1 $ ,Mass $ =\dfrac{1}{2000}\text{ th} $ of a proton and spin $ =\dfrac{1}{2} $

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