# Question #9d679

Apr 13, 2016

${\text{_95^246Am ➝ ""_93^242Np + }}_{2}^{4} \alpha$

#### Explanation:

An alpha particle has a proton number of 2 and mass number of 4. So we need to subtract 2 from Americium's proton number, that will be the proton number for the daughter nucleus:
95 – 2 = 93

Now look up the element with that proton number on the periodic table. Neptunium has a proton number of 93. The mass number of the isotope is calculated by subtracting 4 from Americium's mass number:
246 – 4 = 242

Now we can write the nuclear equation.

${\text{_95^246Am ➝ ""_93^242Np + }}_{2}^{4} \alpha$

Note that the mass and proton numbers on both sides of the equation balance.