Question

The purpose of producing 99Mo (usually by neutron activation of natural molybdenum, as in the preceding problem) is to produce 99mTc. Using the rules, verify that the β− decay of 99Mo produces 99mTc. (Most 99mTc nuclei produced in this decay are left in a metastable excited state denoted 99mTc.)

a.True
b.False

Answer 1

The statement that 99Mo decays into 99mTc is true. This process occurs via beta-minus decay, resulting in a metastable state of technetium, which is heavily utilized in medical imaging due to its short half-life and clear gamma ray emission.

Step-by-step explanation:

Is the statement that the decay of 99Mo produces 99mTc true or false? This statement is true. The process involves the beta-minus decay (β--decay) of 99Mo to produce the metastable state of the technetium isotope, 99mTc.

During β- decay, a neutron inside the molybdenum nucleus is transformed into a proton, and the nucleus emits an electron (beta particle) and an antineutrino. This increases the atomic number by 1, resulting in the formation of technetium (Tc) while the atomic mass number remains unchanged at 99.

The resultant technetium atom is most often left in an excited metastable state, referred to as 99mTc, which subsequently emits a gamma ray (γ ray) to transition to the ground state. The technetium-99m produced is a radiopharmaceutical used extensively in medical imaging. The reaction for the transformation is:

99Mo → 99mTc + β- + v

The metastable state (m in 99mTc) signifies that the nucleus has higher energy than the ground state, and the difference in energy is released in the form of a gamma ray when the nucleus transitions to the ground state.

The use of 99mTc in medical procedures is due to its short half-life (about 6 hours), which minimizes radiation exposure while still permitting high-quality diagnostic imaging.