Question

I need balanced equations written for decays of multiple elements. Please help!alpha decay of Pu-239

Answer 1

An alpha decay happens when an alpha particle is released from a nucleus.

An alpha particle is the helium atom with two neutrons. Since the helium have atomic number 2, than the mass number of an alpha particle is 4, so the alpha particle is:

$^2_4He$

The atom Pu-239 is plutonium, and its atomic number is 94. The given number is the mass number, so its symbol is:

$^{94}_{239}Pu$

So, the reaction is:

$^{239}_{94}Pu\to^4_2He+^m_zX$

We need to figure z, m and X out.

Since the atomic number or the alpha particle is 2, it has 2 protons, so two of the 94 original protons are release, and 92 are left. Thus:

$\begin{gathered} z=92 \\ X=U \end{gathered}$

The mass number of the alpha particle is 4, so form the 239 original mass number, we are left with 235, so:

$m=235$

Thus, the balanced equation is:

$^{239}_{94}Pu\to^4_2He+^{235}_{92}U$

The alpha particle can also use the symble α instead of He, so it would be equivalent.