Question

Identify the following substances involved in radioactive reactions, describing how you obtained the answer:

a. An element which undergoes alpha decay to form uranium-236
b. The product of the alpha decay of americium-240
c. The product of beta decay of tellurium-128

Answer 1

The product of the alpha decay of americium-240 is neptunium-236, and the product of the beta decay of tellurium-128 is iodine-128.

Step-by-step explanation:

The product of the alpha decay of americium-240 (Am-240) is an element that has two fewer protons and an atomic mass that is four units less, because an alpha particle consists of 2 protons and 2 neutrons. Thus, subtracting 2 protons from americium (which has the atomic number 95), we obtain atomic number 93, which is neptunium (Np). Similarly, subtracting 4 from the atomic mass (240) gives us a mass number of 236. Therefore, the equation for the alpha decay of Am-240 is:

^{240}_{95}Am \rightarrow ^{236}_{93}Np + ^{4}_{2}He

The product of the beta decay of tellurium-128 (Te-128) involves the conversion of a neutron into a proton, which increases the atomic number by 1, but leaves the atomic mass unchanged. This produces iodine-128 (I-128) because iodine has an atomic number one higher than tellurium. The balanced nuclear equation is:

^{128}_{52}Te \rightarrow ^{128}_{53}I + ^{0}_{-1}e (beta particle)