109k views
0 votes
A radioisotope decays to give an alpha particle and

Rn-222. What was the original isotope?
a) Uranium-238
b) Polonium-214
c) Radium-226
d) Thorium-232

User Sfufoet
by
7.6k points

1 Answer

3 votes

Final answer:

The original isotope that decays to give an alpha particle and Rn-222 is Radium-226 (Ra-226). This is determined by adding the mass and atomic numbers of the alpha particle (4 and 2) to those of Rn-222, which leads to a mass number of 226 and an atomic number of 88, identifying Ra-226 as the original isotope.

Step-by-step explanation:

To determine the original isotope that decays to give an alpha particle and Rn-222, we need to consider the emissions involved in the decay process. Alpha decay involves emitting an alpha particle from the nucleus, which is a helium nucleus with two protons and two neutrons. Since an alpha particle is represented by 4 in mass number and 2 in atomic number, we subtract these from the mass number and atomic number of Rn-222 to find the original isotope.

Radon-222 has an atomic number of 86 and a mass number of 222. Subtracting the alpha particle properties (mass number = 4, atomic number = 2), we find that the original isotope must have had an atomic number of 86 + 2 = 88 and a mass number of 222 + 4 = 226. Hence, the original isotope is Radium-226 (Ra-226).

Ra-226 undergoes alpha decay to form Rn-222 plus an alpha particle, which can be written in the nuclear equation as:

^(226)_(88)Ra → ^(222)_(86)Rn + ^4_2He

User Hasina Ansari
by
7.1k points