Nuclear disarmament could be accomplished if weapons were not "replenished." The tritium in warheads decays to helium with a half-life of 12.26 yr and must be replaced or the weapon is useless. What fraction of the tritium is lost in 5.50 yr?

Radioactive decay is a natural process by which an unstable atomic nucleus loses energy by emitting particles or electromagnetic waves. This process changes the nucleus into a different element or a different isotope of the same element. Different radioactive isotopes decay at different rates, characterized by their decay constants and half-lives. Decay can involve the emission of alpha particles (helium nuclei), beta particles (electrons or positrons), or gamma rays (high-energy photons). Over time, a radioactive substance will transform until it becomes stable, although the time scales for this can range from fractions of a second to millions of years.

The half-life is the time it takes for half of the material in a sample to decay. It is a constant value specific to each radioactive isotope. To calculate the half-life, we use the formula: \ \ \( t_{1/2} = \frac{\ln{2}}{\lambda} \)

\ Here, \ \( \ln{2} \) is the natural logarithm of 2 (approximately 0.693), and \ \( \lambda \) \ is the decay constant. Once we know the half-life, we can calculate the decay constant with: \ \( \lambda = \frac{\ln{2}}{t_{1/2}} \) \ In the example of tritium decaying into helium, the half-life is 12.26 years. By plugging this value into the decay constant formula, we get: \ \( \lambda = \frac{\ln{2}}{12.26} \approx 0.0565 \text{ yr}^{-1} \)

This decay constant tells us the fraction of tritium that decays per year.

The decay constant, often denoted by the Greek letter \ ( \( \lambda \) ) \ , is a measure of the probability of decay of a radioactive atom per unit time. It provides a link between the half-life and the exponential nature of radioactive decay. A higher decay constant means the substance decays more quickly. Using the formula: \ \( N(t) = N_0 e^{\lambda t} \),

we can determine the amount of substance remaining after a given period.

- \ ( \( N(t) \) ) \ is the quantity of substance left after time \ ( \( t \) ) \.
- \ ( \( N_0 \) ) \ is the initial quantity of the substance.
For the tritium example in the original problem:
\ ( \( N(5.50) = N_0 e^{-\lambda \times 5.50} \approx N_0 e^{-0.31075} \approx 0.732N_0 \) )

This tells us that after 5.50 years, approximately 73.2% of the initial tritium remains.
By subtracting this from 1, we find the lost fraction: \ ( \( 1 - 0.732 = 0.268 \) )\ Thus, 26.8% of the tritium is lost after 5.50 years.