Question

Problem #1 : Radioisotopes are used for a variety of medical purposes including as cancer treatments. Attaching an alpha emitter to an antibody allows for a targeted therapy which can be less damaging to healthy tissue than some other cancer treatments. One such alpha emitter is Astatine 211 (211At) which undergoes LaTeX: \alpha α decay (hence the name, alpha emitter).

1.) What isotope is created by the LaTeX: \alpha α decay of 211At? 2.) How many p+, n0, and e- does it have?

Problem #2 : The grand dream of the protoscience called Alchemy was the transmutation of Lead (Pb) into Gold (Au). Thousands of years later, surely we can do better... after all, Platinum (Pt) is what everyone wants these days. We'll start with an isotope of lead, 209Pb which undergoes LaTeX: \beta β - decay. In reality, that is the end of our tale. However, this is a thought exercise and we can do what we want. Using the different types of radioactive decay described above, but still starting with LaTeX: \beta β - decay, how could we get 209Pb to Pt?

1.) What isotope is created by the LaTeX: \beta β - decay of 209Pb?
2.) Come up with a sequence of radioactive decay steps that would then lead to an isotope of Pt.
3.) What isotope of Pt did you create? How many p+, n0, and e- does it have?

Answer 1

Answer and Explanation: Many nuclei are radioactive, which means they emit particles to become stable. In the process, they also become a different element. There are 3 types of decay:

• Alpha decay: it emits a particle of Helium, i.e., emits a particle with 2 as atomic number (Z) and 4 as atomic mass (A);

• Beta decay: emits an electron: a particle with 0 mass and -1 as atomic number;

• Gamma decay: emits a high energy form of electromagnetic radiation and it is extremely dangerous and penetrating;

Problem #1:

1.) Astatine: Z = 85 and A = 211

Alpha decay:
`{{A=211} \atop {Z=85}} \right. At` ⇒
`{{207} \atop {83}} \right. Bi + {{4} \atop {2}} \right. \alpha`

The isotope created is Bismuth

Characteristics: Z = 83; e⁻ = 83; n = A - Z = 207 - 83 = 124

The isotope is Bismuth with 83 protons, 83 electrons and 124 neutrons.

Problem #2

1.) Lead: Z = 82 and A = 82

Beta decay:
`{{209} \atop {82}} \right. Pb` ⇒
`{{209} \atop {81}} \right. Tl+{{0} \atop {-1}} \right. \beta`

The isotope created is talium.

2.)
`{{209} \atop {82}} \right. Pb` ⇒
`{{209} \atop {81}} \right. Tl+{{0} \atop {-1}} \right. \beta`

`{{209} \atop {81}} \right. Tl` ⇒
`{{205} \atop {79}} \right. Au+{{4} \atop {2}} \right. \alpha`

`{{205} \atop {79}} \right. Au` ⇒
`{{205} \atop {78}} \right. Pt+{{0} \atop {-1}} \right. \beta`

3.) The isotope created is
`{{205} \atop {78}} \right. Pt`.

p⁺ = 78; e⁻ = 78; n = 127

The isotope created ahs 78 protons, 78 electrons and 127 neutrons.