Question

(a) What do you mean by $Q$ value of a nuclear reaction?
(b) Write down the expression for $Q$ value in the class of $\alpha$ decay.

Answer 1

$Q$ value can be calculated from the reactant masses and the products. $Q$ value in $\alpha$ decay is the variation between the parent's mass and the daughter's coupled mass and the $\alpha$-particle, multiplied by ${c^2}$.

Complete step by step solution:
(a) Nuclear reaction, alteration of an atomic nucleus' personality, or properties, caused by bombarding it with an energetic particle. The particle that is bombarding may be an alpha particle, a gamma-ray electron, a neutron, a proton, or a heavy ion.
The $Q$-value for a reaction in nuclear physics and chemistry is the amount of energy consumed or emitted during the nuclear reaction. The importance is related to the enthalpy of a chemical reaction or the radiation of products of radioactive decay. It can be calculated from the reactant masses and the products.
The value is given by,
$Q = ({m_r} - {m_p}) \times 931\;{\text{MeV}}$
Where weights are in units of atomic mass. Both ${m_r}$ and ${m_p}$ are respectively the sums of the masses of reactants and of products.

(b) $\alpha$ - Decay: $\alpha$ - Decay is the $\alpha$ -particle nuclear emission and is ${}_2^4{\text{He}}$, two protons and two neutrons. This is a very stable nucleus, since magic is equal. The nucleus for the daughter has two protons and four nucleons less than the nucleus for the parent.

$Q\alpha$ is the variation between the parent's mass and the daughter's coupled mass and the $\alpha$-particle, multiplied by ${c^2}$, that is
$Q\alpha = \left( {{m_P} - {m_D} - {m_\alpha }} \right){c^2}$. Generally the Liquid Drop Model can estimate the mass difference between both the parent and daughter nucleus fairly well.

It is equivalent to the difference between the sum of the daughter's binding energies and the $\alpha$-particles, and the parent nucleus.

Note: The importance is related to the enthalpy of a chemical reaction or the radiation of products of radioactive decay. Generally, the higher the positive $Q$ value for the reaction, the quicker the reaction and the more likely the reaction is to "favour" the goods.