Final answer:
The rest of the x-ray energy that is not stored in the phosphor layer of the image plate is lost as heat or other forms of energy that are not utilized in the imaging process.
Step-by-step explanation:
When approximately half of the x-ray energy is stored in the phosphor layer of the image plate during radiographic imaging, the other half is not stored and is ultimately lost in various forms. The phosphor layer absorbs the x-ray photons and stores part of their energy. The rest of the energy is not captured efficiently, and can be dissipated as low-level heat or other forms of energy that are not utilized in the imaging process. When the stored x-ray energy is read using a red laser, the phosphor layer emits this stored energy as visible light, which is then captured to form an image. However, this visible light emission represents only the energy that was successfully stored, not the entirety of the initial x-ray energy.
It is helpful to understand that in various materials, when electrons absorb energy, not all of this energy is emitted as visible light due to losses such as heat during the process. Regarding the options provided: (a) red laser light is used to read the stored image, not what happens to the unused energy; (b) 'linear' is not an appropriate term in this context; (c) is related to the storage medium, which is correct, but does not represent what happens to the rest of the x-ray energy; (d) some energy is indeed emitted as visible light, but only after stimulation by the red laser light during the readout process, not at the time of x-ray exposure. The correct answer to what happens to the other half of the x-ray energy is that it could be lost as heat or other non-useful forms of energy transfer.