\(^{226}_{88}Ra\) emits two alpha particles to produce polonium(po) nuclide. The correct formula for the nuclide?
\(^{224}_{86}po\)
\(^{222}_{86}po\)
\(^{222}_{84}po\)
\(^{218}_{84}po\)
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To determine the correct formula for the nuclide produced when 226^88Ra emits two alpha particles, we need to understand the process of alpha decay and the resulting nuclide.
Alpha decay involves the emission of an alpha particle, which is a helium-4 nucleus (
4He2 consisting of two protons and two neutrons. When an alpha particle is emitted, the atomic number of the parent nuclide decreases by 2, and the mass number decreases by 4.
In this case,
226
Ra
88 emits two alpha particles. Therefore, the atomic number decreases by
2×2=4, and the mass number decreases by 4×2=8.
The atomic number of polonium (Po) is 84, and its most common isotope has a mass number of 210. Therefore, the correct formula for the nuclide produced would be
226 224
Ra → Po
88 84
So, the correct answer is:
C.
222
Po
84
