Question

Until roughly 1970, tritium (3H), a radioactive isotope of hydrogen, was a component of fluorescent watch dials and hands. For 3H, t1/2 = 12.3 yr. Assume you have such a watch. If a minimum of 12.0% of the original tritium is needed to read the dial in dark places, for how many years could you read the time at night? Assume first-order kinetics.

Answer 1

Given,

t_1/2 of the tritium(³H) = 12.3 years

Assuming,

the initial amount of tritium in the watch (N_o) = 100 %

The amount of tritium remaining after time "t" (N_t) = 12.0 %

Firstly calculating the decay constant() from the given half-life period of ³H,

= 0.693 / t_1/2

= 0.693 /12.3 yr

= 0.05634 yr^-1

We know, the first order equation,

ln [N_t / N_o] = - x t

Substituting the known values,

ln [12 / 100] = - 0.05634 yr^-1 x t

t = 37.6 years