What is the product of alpha decay of curium-226? What product is formed when curium-226 undergoes alpha decay?


Curium has atomic number 96. Since alpha decay decreases the number of protons and neutrons by 2, the product formed has atomic number 94 and atomic mass 222, thus the product is plutonium-222. However this was not an answer choice. The only answer choices were isotopes of radon, radium, and thorium. Either I don't know how to do a very simple question, or there are alternative representations of radioactive isotopes that I don't know about.

I would like to determine, if the question itself is wrong or if I am making a mistake.
The question is flawed (possibly a misprint), your reasoning is sound (given the information provided to you).
Curium-226 does not exist, according to the Web Elements page on the isotopes of curium, the 'smallest' isotope is curium-240, which undergoes $\alpha$ decay to become plutonium-236.
$$\ce{^{240}_{96}Cm -&gt; ^{236}_{94}Pu + ^{4}_{2}\alpha}$$
The most likely candidate of what the question should have been asking about is radium-226 decaying to radon-222,
$$\ce{^{226}_{88}Ra -&gt; ^{222}_{86}Rn + ^{4}_{2}\alpha}$$
which is part of the uranium-series decay chain:

This is the most likely scenario, as the smallest isotope of thorium is thorium-227 (WebElements) and for radium, it is radium-223 (WebElements), both have too high a mass for anything with an atomic mass of 226 to directly decay into by alpha decay (as per the question).