1. Write balanced equations for each of the following processes.
a. Chromium- 51, which targets the spleen and is used as a tracer in studies of red blood cells, decays by electron capture.
b. Iodine-131, used to treat hyperactive thyroid glands, decays by producing a beta particle.
2. In each of the following nuclear reactions, supply the missing particle?
a. ^60Co -> ^60Ni + ?
b. ^97Tc + ? -> ^97Mo
c. ^99Tc -> ^99Ru + ?
d. ^239Pu -> ^235U + ?
3. A chemist wishing to do an experiment requiring 47Ca (half life = 4.5 days) needs 5.0ìg of the
nuclide. What mass of 47Ca must be ordered if it takes 48 h for delivery from the supplier? Assume that the atomic mass of ^47Ca iis 47.0
4. What is the ratio of ^209Pb/^238U by mass in a rock that is 4.5x10^9 years old? (For 238 U, t1/2= 4.5x10^9 years)
Please show work.
1a. The process is chromium-51 decaying by electron capture. Electron capture is when an electron from the inner shell of an atom is captured by the nucleus, causing a proton to convert into a neutron. Therefore, the balanced equation for this process is:
^51Cr + e- -> ^51V
b. The process is iodine-131 decaying by producing a beta particle. A beta particle is an electron or a positron emitted from the nucleus. Therefore, the balanced equation for this process is:
^131I -> ^131Xe + e-
2a. In this nuclear reaction, the missing particle after ^60Co decays into ^60Ni is a beta particle (e-):
^60Co -> ^60Ni + e-
b. In this nuclear reaction, the missing particle before ^97Tc transforms into ^97Mo is a beta particle (e-):
^97Tc + e- -> ^97Mo
c. In this nuclear reaction, the missing particle after ^99Tc decays into ^99Ru is a beta particle (e-):
^99Tc -> ^99Ru + e-
d. In this nuclear reaction, the missing particle before ^239Pu transforms into ^235U is an alpha particle (α):
^239Pu -> ^235U + α
3. To determine the mass of 47Ca needed for the experiment, we first need to calculate the number of radioactive decay events that occur during the 48-hour delivery time. We can use the radioactive decay equation:
N = N0 * (1/2)^(t/T)
Here, N is the final number of radioactive atoms, N0 is the initial number of radioactive atoms, t is the time, and T is the half-life.
Given:
- Half-life of ^47Ca = 4.5 days (converted to 4.5 * 24 = 108 hours)
- Initial mass of 47Ca needed = 5.0μg
Using the atomic mass of ^47Ca (47.0 g/mol), we can convert μg to grams:
Initial mass in grams = (5.0 * 10^-6 g) / (1 * 10^6) = 5.0 * 10^-12 g
Now, we can calculate the initial number of radioactive atoms (N0) using Avogadro's number (6.022 * 10^23 atoms/mol):
Initial number of atoms (N0) = (5.0 * 10^-12 g) / (47.0 g/mol) * (6.022 * 10^23 atoms/mol) = 6.436 * 10^9 atoms
Next, we need to calculate the final number of atoms (N) after 48 hours:
N = N0 * (1/2)^(t/T) = (6.436 * 10^9) * (1/2)^(48/108) ≈ 2.281 * 10^9 atoms
Finally, to determine the mass of 47Ca needed, we can multiply the final number of atoms (N) by the atomic mass of ^47Ca:
Mass of 47Ca needed = (2.281 * 10^9 atoms) * (47.0 g/mol) = 1.072 * 10^11 g ≈ 107.2 g
Therefore, approximately 107.2 grams of ^47Ca must be ordered for the experiment.
4. The ratio of ^209Pb/^238U by mass in a rock that is 4.5x10^9 years old can be determined using the concept of radioactive decay and the decay constant (λ). The decay constant can be calculated using the half-life (Ƭ):
λ = ln(2) / Ƭ
Given:
- Half-life of ^238U = 4.5x10^9 years
Using the decay constant (λ) formula, we find:
λ = ln(2) / (4.5x10^9) = 0.154 x 10^-9 year^-1
Now, we need to calculate the ratio of ^209Pb to ^238U after 4.5x10^9 years. Let's assume we start with 1 gram of ^238U:
Amount of ^238U remaining after 4.5x10^9 years = (1 gram) * e^(-λ * 4.5x10^9) = (1 gram) * e^(-0.154) ≈ 0.857 grams
Amount of ^209Pb produced = 1 gram - 0.857 grams = 0.143 grams
Therefore, the ratio of ^209Pb/^238U by mass in the 4.5x10^9-year-old rock is:
0.143 grams / 1 gram = 0.143