Question

In each of the parts of this question, a nucleus undergoes a nuclear decay. Determine the resulting nucleus in each case.

Part A

227/89 Ac undergoes alpha decay. Determine the resulting nucleus.

For example, if the resulting nucleus is 40/20 Ca enter ^40_20Ca.

Part B

211/83 Bi undergoes beta-minus decay. Determine the resulting nucleus.

For example, if the resulting nucleus is 40/20 Ca enter ^40_20Ca.

Part C

22/11 Na undergoes beta-plus decay. Determine the resulting nucleus.

For example, if the resulting nucleus is 40/20 Ca enter ^40_20Ca.

Part D

98/43 Tc undergoes gamma decay. Determine the resulting nucleus.

Enter your answer in the space below. For example, if the resulting nucleus is 40/20 Ca enter ^40_20Ca.

Answer 1

Part A )
During alpha decay, an atom's nucleus sheds 2 protons and 2 neutrons.
As a result the remaining nucleus has an atomic mass number reduced by 4 (227-4 = 223), and an atomic/proton number reduced by 2 (89-2 = 87) ..

227/89 Ac → 223/87 X
X = Fr (Francium)

The resulting nucleus is Francium 223/87 Fr
Part B)

During β- decay a neutron decays to a proton.
As a result the remaining nucleus has not changed it's atomic mass number (same number of particles in the nucleus), but it's atomic/proton number has increased by 1.

Therefore -
211/83 Bi → 211/84 X
X = Po (Polonium)

The resulting nucleus is Polonium 211/84 Po

Part C)
During β+ a proton is converted to a neutron ..
As a result the remaining nucleus has not changed it's atomic mass number (same number of particles in the nucleus), but it's atomic/proton number has decreased by 1.

Therefore -
22/11 Na → 22/10 X
X = Ne (Neon)

The resulting nucleus is Neon 22/10 Ne

Part D)
During a Gamma Decay ,the number of protons (and neutrons) in the nucleus does not change in this process, so the parent and daughter atoms are the same chemical element.

Therefore the resulting nucleus is same 98/43 Tc.