The isotope \({}^{208}{\rm{Tl}}\) undergoes \({\rm{\beta }}\) decay with a half-life of \({\rm{3}}{\rm{.1 min}}\).

(a) What isotope is produced by the decay?

(b) How long will it take for \({\rm{99}}{\rm{.0\% }}\) of a sample of pure \({}^{208}{\rm{Tl}}\) to decay?

(c) What percentage of a sample of pure \({}^{208}{\rm{Tl}}\) remains un-decayed after \({\rm{1}}{\rm{.0 h}}\)?

In nuclear physics, beta decay (\({\rm{\beta - }}\)decay) is a type of radioactive decay in which an atomic nucleus emits a beta particle (fast, energetic electron or positron) that transforms the original nuclide into an isobar of that nuclide.

The formula to calculate the ratio of the radioactive substance:

\(\frac{{\rm{N}}}{{{{\rm{N}}_{\rm{0}}}}}{\rm{ = }}{{\rm{e}}^{{\rm{ - \lambda }} \cdot {\rm{t}}}}\)

Where\({\rm{t}}\)is time,\({\rm{\lambda }}\)is decay constant,\({{\rm{N}}_{\rm{0}}}\)is initial concentration, and\({\rm{N}}\)is the concentration at time\({\rm{t}}\).

The beta decay is represented as\({}_{{\rm{ - 1}}}^{\rm{0}}{\rm{e}}\).

That means the atomic number increases by\({\rm{1}}\), and the mass number remains the same.

Now,\({}_{{\rm{81}}}^{{\rm{208}}}{\rm{Tl}}\)and if the atomic number increases by\({\rm{1}}\), which is\({\rm{82}}\), the element in the periodic table which has atomic number\({\rm{82}}\)is\({\rm{Pb}}\).

Therefore, the isotope is \({}_{82}^{208}{\rm{Pb}}\).

From the half-life, calculate\({\rm{\lambda }}\):

\(\begin{array}{l}{\rm{\lambda = }}\frac{{{\rm{ln(2)}}}}{{{{\rm{t}}_{{\rm{1/2}}}}}}\\{\rm{\lambda = }}\frac{{{\rm{ln(2)}}}}{{{\rm{3}}{\rm{.1 min}}}}\\{\rm{\lambda = 0}}{\rm{.22 mi}}{{\rm{n}}^{{\rm{ - 1}}}}\end{array}\)

The\({\rm{99\% }}\)got decayed, which means the remaining amount is\({\rm{1\% }}\).

\(\begin{array}{c}\frac{{\rm{N}}}{{{{\rm{N}}_{\rm{0}}}}}{\rm{ = }}\frac{{\rm{1}}}{{{\rm{100}}}}\\\frac{{\rm{N}}}{{{{\rm{N}}_{\rm{0}}}}}{\rm{ = 0}}{\rm{.01}}\end{array}\)

Rearranging and solving the equation:

\(\begin{array}{c}\frac{{\rm{N}}}{{{{\rm{N}}_{\rm{0}}}}}{\rm{ = }}{{\rm{e}}^{{\rm{ - \lambda }} \cdot {\rm{t}}}}\\{\rm{0}}{\rm{.01 = }}{{\rm{e}}^{{\rm{ - 0}}{\rm{.22mi}}{{\rm{n}}^{{\rm{ - 1}}}} \cdot {\rm{t}}}}\\{\rm{ln(0}}{\rm{.01) = }}\left( {{\rm{ - 0}}{\rm{.22\;mi}}{{\rm{n}}^{{\rm{ - 1}}}} \cdot {\rm{t}}} \right)\\{\rm{t = }}\frac{{{\rm{ - ln(0}}{\rm{.01)}}}}{{{\rm{ - 0}}{\rm{.22\;mi}}{{\rm{n}}^{{\rm{ - 1}}}}}}{\rm{t}}\\{\rm{ = 20}}{\rm{.9\;min}}\end{array}\)

Therefore, the value for time is obtained as \({\rm{20}}{\rm{.9\;min}}\).

The time given is \({\rm{1 hr = 60 min}}\).

The ratio of \({{\rm{N}}_{\rm{0}}}\) can be calculated as follows:

\(\begin{array}{c}\frac{{\rm{N}}}{{{{\rm{N}}_{\rm{0}}}}}{\rm{ = }}{{\rm{e}}^{{\rm{ - 0}}{\rm{.22\;mi}}{{\rm{n}}^{{\rm{ - 1}}}} \cdot {\rm{60\;min}}}}\\\frac{{\rm{N}}}{{{{\rm{N}}_{\rm{0}}}}}{\rm{ = 1}}{\rm{.8}} \cdot {\rm{1}}{{\rm{0}}^{{\rm{ - 6}}}} \cdot {\rm{100}}\\\frac{{\rm{N}}}{{{{\rm{N}}_{\rm{0}}}}}{\rm{ = 1}}{\rm{.8}} \cdot {\rm{1}}{{\rm{0}}^{{\rm{ - 4}}}}{\rm{\% }}\end{array}\)

Therefore, the value for the ratio is obtained as \({\rm{1}}{\rm{.8}} \cdot {\rm{1}}{{\rm{0}}^{{\rm{ - 4}}}}{\rm{\% }}\).