Explain Enthalpy of atomization $\left( \Delta_{a} H^{\theta} \right)$.

The enthalpy of atomization $\left( \Delta_{a} H^{\theta} \right)$ is defined as the enthalpy change occurring when one mole of bonds in a compound is completely broken to obtain atoms in the gas phase.
For example,in the case of dihydrogen:
$H_{2(g)} \rightarrow 2H_{(g)} ; \Delta_{a} H^{\theta} = 435.0 \ kJ \ mol^{-1}$
In this process,the $H-H$ bond is broken to form gaseous $H$ atoms. For diatomic molecules,the enthalpy of atomization is equivalent to the bond dissociation enthalpy.
For polyatomic molecules like methane:
$CH_{4(g)} \rightarrow C_{(g)} + 4H_{(g)} ; \Delta_{a} H^{\theta} = 1665 \ kJ \ mol^{-1}$
For solid elements like sodium:
$Na_{(s)} \rightarrow Na_{(g)} ; \Delta_{a} H^{\theta} = 108.4 \ kJ \ mol^{-1}$
In this case,the enthalpy of atomization is equal to the enthalpy of sublimation.