Question

Radiological laboratories often have a container of molybdenum-99, which decays to form technetium-99m. The equation for the decay is the following 9942mo→99m43tc. What other particle is formed?

A. Electron
B. Alpha particle
C. Beta particle
D. Gamma ray

Answer 1

During the decay of molybdenum-99 to technetium-99m, a beta particle, which is an electron, is emitted from the nucleus. This is represented by the increase in atomic number from 42 to 43 in the nuclear equation.

Step-by-step explanation:

The decay of molybdenum-99 (9942Mo) to form technetium-99m (99m43Tc) involves the emission of a particular type of particle. Given the nuclear equation 9942Mo → 99m43Tc, we can determine the missing particle by looking at the atomic numbers and mass numbers before and after the decay. The atomic number increases by 1 (from 42 to 43), indicating the emission of a beta particle, which is an electron (e-) emitted from the nucleus during beta decay.

Therefore, the correct answer to the question 'What other particle is formed during the decay of 9942Mo to 99m43Tc?' is C. Beta particle. The complete nuclear reaction is as follows:

9942Mo → 99m43Tc + e-