Why doesn't the atomic number change by two in the process of production of Americium?

Looking at the bomardment of uranium-238 this is what I see.

First, I see a uranium-238 atom with 92 protons bombarded with a neutron. The neutron then undergoes beta decay into a proton. So we then get Neptunium as the neutron decayed into a proton and hence, atomic and mass number increase by one. An electron is also released from the atom.

That makes sense.

Looking at the second equation of the bombardment of Plutonium-239.

Plutonium-239 with 94 protons gets bombardment with 2 neutrons. These neutrons undergo beta decay to produce protons. Thus, the atomic and mass number must increase by two. The mass number, indeed, does increase by two. The atomic number does not. An electron is also released? Not sure why aswell.

• Only one neutron decays. – Jon Custer Feb 15 '15 at 1:05

What might be confusing is that it's actually two reactions. Two neutrons are captured followed by a single beta decay:

$$\ce{^239_94Pu ->[2(n,\gamma)]^241_94Pu->[\beta^-]^241_95Am}$$

The first reaction leaves the new nucleus in an excited state, so a gamma ray is emitted. The second reaction is actually quite slow with a halflife of ~14 years. It's a beta minus decay so that's why an electron is emitted.