Final answer:
Thorium-231 undergoes alpha decay to form Radium-227 and an alpha particle. If there is also gamma decay, Radium-227 releases a gamma photon, resulting in no change to its mass or atomic number.
Step-by-step explanation:
The question asks for the nuclear equation that represents both the alpha and gamma decay of Thorium-231. To write this, we need to know that during alpha decay, the atom loses an alpha particle (which is equivalent to a helium nucleus), resulting in the atomic number decreasing by 2 and the mass number by 4. Gamma decay, on the other hand, involves the emission of gamma radiation and does not change the atomic number or the mass number of the nucleus.
For Thorium-231 (which is represented as 231Th), alpha decay would be written as follows:
^231_90Th → ^227_88Ra + ^4_2He
This equation shows Thorium-231 decaying to form Radium-227 and an alpha particle. If Thorium-231 also undergoes gamma decay, it would simply release a gamma photon (γ) without altering the mass or atomic numbers of the Radium-227.
^227_88Ra → ^227_88Ra + γ
The above gamma decay equation represents the emission of gamma radiation from Radium-227, which would occur after the alpha decay if Thorium-231 was also experiencing gamma decay. The Radium-227 remains unchanged except for its energy state due to the release of a gamma photon.