Question

Smoke detectors contain a small amount of americium-241. Its decay product is neptunium-237. Identify the emission from americium-241.

Short answer

Answer: The emission resulting from the decay of americium-241 into neptunium-237 is an alpha particle (α).

Step-by-step solution

Determine initial and final atomic and mass numbers for the two isotopes

First, we need to find the atomic (protons) and mass numbers (protons + neutrons) for both americium-241 and neptunium-237. The atomic number (Z) specifies the number of protons in the nucleus, while the mass number (A) is the sum of the number of protons and neutrons. Americium (Am) has an atomic number of 95, so it has 95 protons. Americium-241, therefore, has a mass number of 241 (95 protons + 146 neutrons). Neptunium (Np) has an atomic number of 93, so it has 93 protons. Neptunium-237, therefore, has a mass number of 237 (93 protons + 144 neutrons).

Identify changes in atomic and mass numbers during decay

Now, we will analyze the difference between the initial and final atomic and mass numbers due to the decay of americium-241 into neptunium-237. Change in atomic number: ΔZ = 93 - 95 = -2 Change in mass number: ΔA = 237 - 241 = -4

Determine the type of emission based on changes in atomic and mass numbers

From the changes in atomic and mass numbers, we can determine the type of radioactive decay or the emission released during this process: 1. Alpha decay: In this type of decay, the nucleus releases an alpha particle, which consists of 2 protons and 2 neutrons. Therefore, an alpha decay would have ΔZ = -2 and ΔA = -4. In this case, our changes in atomic and mass numbers match, so the emission must be an alpha particle. Thus, the emission from the decay of americium-241 into neptunium-237 is an alpha particle (α).