In the given reaction, find value of Q value. ₆C13 → ₆C12 + ₀n1 + (Q-value) Given : mass of ₆C¹³ ⇒ x mass of ₆C¹² ⇒ y mass of ₀n¹ ⇒ z

The correct option is (y + z – x) C².

To find the Q-value of the given nuclear reaction, we analyze the mass-energy relation of the process:
${}_6^{13}C\to {}_6^{12}C+{}_0^{1}n+\text{Q-value}$

We are given the following masses:
- Mass of the reactant, ${}_6^{13}C$ = $x$
- Mass of the first product, ${}_6^{12}C$ = $y$
- Mass of the second product, ${}_0^{1}n$ = $z$

The total mass of the products is:
$M_{\text{products}}=y+z$

The total mass of the reactants is:
$M_{\text{reactants}}=x$

The mass difference ($\mathrm{\Delta m}$) between the products and the reactants is:
$\mathrm{\Delta m}=(y+z)-x$

Using Einstein's mass-energy equivalence relation, the energy equivalent of this mass difference (the Q-value) is determined by multiplying the mass difference by the square of the speed of light ($C^2$):
$Q=\mathrm{\Delta m}·C^2$
Substituting the expression for $\mathrm{\Delta m}$:
$Q=(y+z-x)C^2$