$\text{In the following nuclear reaction,}{\mathrm{}}_{\mathrm{ 0}}^{\mathrm{<font\; dir="auto"\; style="vertical-align:\; inherit;"><font\; dir="auto"\; style="vertical-align:\; inherit;"> 1</font></font>}}n+{\mathrm{<font\; dir="auto"\; style="vertical-align:\; inherit;"><font\; dir="auto"\; style="vertical-align:\; inherit;">IN</font></font>}}_{92}^{235}\to {\mathrm{<font\; dir="auto"\; style="vertical-align:\; inherit;"><font\; dir="auto"\; style="vertical-align:\; inherit;">Car</font></font>}}_{54}^{140}<font\; dir="auto"\; style="vertical-align:\; inherit;"><font\; dir="auto"\; style="vertical-align:\; inherit;">+</font></font>{\mathrm{Sr}}_a^b<font\; dir="auto"\; style="vertical-align:\; inherit;"><font\; dir="auto"\; style="vertical-align:\; inherit;">+</font></font>2\left({\mathrm{<font\; dir="auto"\; style="vertical-align:\; inherit;"><font\; dir="auto"\; style="vertical-align:\; inherit;"></font></font>}}_{\mathrm{<font\; dir="auto"\; style="vertical-align:\; inherit;"><font\; dir="auto"\; style="vertical-align:\; inherit;">0</font></font>}}^{\mathrm{<font\; dir="auto"\; style="vertical-align:\; inherit;"><font\; dir="auto"\; style="vertical-align:\; inherit;">1</font></font>}}\mathrm{<font\; dir="auto"\; style="vertical-align:\; inherit;"><font\; dir="auto"\; style="vertical-align:\; inherit;">n</font></font>}\right)\text{we have}$

The correct option is: a = 94, b = 38

Step-by-step Explanation:

In any nuclear reaction, two fundamental conservation laws must be satisfied:
1. Conservation of Mass Number (A): The sum of the mass numbers (superscripts, representing total nucleons) of the reactants must equal the sum of the mass numbers of the products.
2. Conservation of Atomic Number (Z): The sum of the atomic numbers (subscripts, representing nuclear charge) of the reactants must equal the sum of the atomic numbers of the products.

Let the strontium nucleus be represented as having a mass number a and an atomic number b.

1. Applying Conservation of Mass Number:

The total mass number on the reactant (left) side is:

$A_{\text{reactants}}=1+235=236$

The total mass number on the product (right) side (including the two neutrons) is:

$A_{\text{products}}=140+a+2\left(1\right)=142+a$

Setting the reactant and product mass numbers equal:

$142+a=236$

Solving for a:

$a=236-142=94$

2. Applying Conservation of Atomic Number (Charge):

The total atomic number on the reactant (left) side is:

$Z_{\text{reactants}}=0+92=92$

The total atomic number on the product (right) side (including the two neutrons) is:

$Z_{\text{products}}=54+b+2\left(0\right)=54+b$

Setting the reactant and product atomic numbers equal:

$54+b=92$

Solving for b:

$b=92-54=38$

Therefore, we have a = 94 (mass number of Sr) and b = 38 (atomic number of Sr).