Radon-222 undergoes alpha decay.what is the nuclear equation
86Rn222 = 2He4 + 84Po218
First number just before Rn is atomic number; number following symbol is mass number.
86Rn222 ==> 2He4 + nXm
We we need to evaluate for n and m. To do that we remember that the atomic numbers must add on both sides and the mass numbers must add up on both sides.
86 = 2+n; therefore, n = 84
222 = 4 + m; therefore, m = 218
Now we look on the periodic table and find element number 84. That is Po; therefore, element X is Po so the equation in full is
86Rn222 = 2He4 + 84Po218
Well, aren't you in for a radioactive treat? The alpha decay of Radon-222 (Rn-222) can be represented by the following nuclear equation:
222Rn ---> 218Po + 4He
In other words, Radon-222 (222Rn) decays and turns into Polonium-218 (218Po), while simultaneously releasing an alpha particle (4He) as a byproduct. Don't worry, though, these alpha particles are just helium nuclei, so it's not that scary!
The nuclear equation for the alpha decay of Radon-222 can be represented as:
^222Rn -> ^218Po + ^4He
In this equation, Radon-222 undergoes alpha decay, resulting in the formation of Polonium-218 (^218Po) and the emission of an alpha particle (^4He).
To determine the nuclear equation for the alpha decay of radon-222, we need to understand that during alpha decay, an alpha particle is emitted from the nucleus.
The alpha particle consists of two protons and two neutrons, which is essentially the same as a helium nucleus. Therefore, the nuclear equation for the alpha decay of radon-222 can be written as follows:
^222Rn --> ^218Po + ^4He
This equation indicates that one atom of radon-222 decays to form one atom of polonium-218 and one alpha particle (helium-4).