Question

The radioactive hydrogen isotope ³H, called tritium, has a

half-life of 12 years.

a. What are the decay mode and the daughter nucleus of tritium?

b. What are the lifetime and the decay rate of tritium?

Short answer

Hence,

$$\begin{gathered} \text{a.)}{}^3\mathrm{H}\to {}^3\mathrm{He}+{\beta }^{-} \\ \text{b.)}\tau =5.46\times {10}^8\mathrm{s} \end{gathered}$$

$$\begin{gathered} \text{a.)}{}^3\mathrm{H}\to {}^3\mathrm{He}+{\beta }^{-} \\ \text{b.)}\tau =5.46\times {10}^8\mathrm{s} \end{gathered}$$

Step-by-step solution

Given information

The radioactive hydrogen isotope³H, called tritium, has a

half-life of 12 years.

Explanation

a.) We already know that the ⁺H is tritium, and that tritium has N = 2 and Z = 1. *H comes the creation daughter nucleus ³He after it has passed beta-minus decay.

${}^3\mathrm{H}\to {}^3\mathrm{He}+{\beta }^{-}$

b.) We must first determine the value of the decay rate.

$$\begin{gathered} r=\frac{1}{\tau } \\ r=\frac{\ln \left(2\right)}{t_{1/2}} \\ r=\frac{0.693}{12\mathrm{yr}} \\ r=\frac{0.693}{12\mathrm{yr}}\times \frac{1\mathrm{yr}}{3.15\times {10}^7\mathrm{s}} \\ r=1.83\times {10}^{-9}\mathrm{s} \end{gathered}$$

Next, we must determine the value of the time constant $\tau$:

$$\begin{gathered} r=\frac{1}{\tau } \\ \tau ·r=1 \\ \tau =\frac{1}{r} \\ \tau =\frac{1}{1.83\times {10}^{-9}\mathrm{s}} \\ \tau =5.46\times {10}^8\mathrm{s} \end{gathered}$$