Question

Iodine that enters the body is stored in the thyroid gland from which it is released to control growth and metabolism. The thyroid can be imaged if iodine\({\rm{ - 131}}\)is injected into the body. In larger doses, I\({\rm{ - 133}}\)is also used as a means of treating cancer of the thyroid. I\({\rm{ - 131}}\)has a half-life of\({\rm{8}}{\rm{.70}}\)days and decays by\({\rm{\beta - }}\)emission. (a) Write an equation for the decay. (b) How long will it take for\({\rm{95}}{\rm{.0\% }}\)of a dose of I\({\rm{ - 131}}\)to decay?

Short answer

1. The equation for decay is obtained as:\(_{{\rm{53}}}^{{\rm{133}}}{\rm{I}} \to _{{\rm{ - 1}}}^{\rm{0}}{\rm{e + }}_{{\rm{53}}}^{{\rm{133}}}{\rm{Xe}}\).
2. It will take \({\rm{37}}{\rm{.6}}\) days to decay.

Step-by-step solution

Step 1: Nuclear Chemistry

Nuclear chemistry is a branch of chemistry that studies radioactivity, nuclear processes, and atomic nuclei alterations such as nuclear transmutation and nuclear characteristics.

Subpart a)

The equation we have is:

\(_{{\rm{53}}}^{{\rm{133}}}{\rm{I}} \to _{{\rm{ - 1}}}^{\rm{0}}{\rm{e + }}_{\rm{y}}^{\rm{x}}{\rm{?}}\)

The mass on the left side is \({\rm{133}}\), which means we also require the total of \({\rm{133}}\) on the right side:

\(\begin{array}{c}{\rm{x + 0 = 133}}\\{\rm{x = 133}}\end{array}\)

Because the atomic number on the left is \({\rm{53}}\), we must also have the total of \({\rm{53}}\) on the right:

\(\begin{array}{c}{\rm{y + ( - 1) = 53}}\\{\rm{y = 54}}\end{array}\)

The element xenon, with an atomic number of \({\rm{54}}\), is found in the periodic table. The reaction is as follows:

\(_{{\rm{53}}}^{{\rm{133}}}{\rm{I}} \to _{{\rm{ - 1}}}^{\rm{0}}{\rm{e + }}_{{\rm{53}}}^{{\rm{133}}}{\rm{Xe}}\)

Therefore, the equation is: \(_{{\rm{53}}}^{{\rm{133}}}{\rm{I}} \to _{{\rm{ - 1}}}^{\rm{0}}{\rm{e + }}_{{\rm{53}}}^{{\rm{133}}}{\rm{Xe}}\).

Subpart b)

There is a simple formula for calculating the time of decay:

The value of \({\rm{t}}\) is time, \({\rm{\lambda }}\) is the radioactive decay constant, \({{\rm{n}}_{\rm{0}}}\) is the starting concentration of the radioactive material, and \({{\rm{n}}_{\rm{t}}}\) is the concentration of the radioactive substance over time \({\rm{t}}\).

We may deduce \({\rm{\lambda }}\) from the half-life:

\(\begin{array}{c}{\rm{\lambda = }}\frac{{{\rm{In(2)}}}}{{{{\rm{t}}_{{\rm{1/2}}}}}}\\{\rm{ = }}\frac{{{\rm{In(2)}}}}{{{\rm{8}}{\rm{.70 days}}}}\\{\rm{ = 0}}{\rm{.07967 days}}\end{array}\)

Then the value of\({{\rm{n}}_{\rm{0}}}\)is\(1\)and\({{\rm{n}}_{\rm{t}}}\)is of\({\rm{95\% }}\).

So, we get:

\(\begin{array}{c}{{\rm{n}}_{\rm{t}}}{\rm{ = 0}}{\rm{.95 \times 1}}\\{\rm{ = 0}}{\rm{.05}}\end{array}\)

The time is then obtained as:

\(\begin{array}{c}{\rm{t = }}\frac{{\rm{1}}}{{{\rm{0}}{\rm{.07967 day}}}}{\rm{In}}\frac{{\rm{1}}}{{{\rm{0}}{\rm{.05}}}}\\{\rm{ = 37}}{\rm{.6 day}}\end{array}\)

Therefore, the decay time will be: \({\rm{37}}{\rm{.6 day}}\).