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 14.0% 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

`t ~~ 34.889" yr"`

Explanation:

Given `t_(1/2) = 12.3" yr"`

That amount at any given time, `A(t)`, is a power function with `1/2` as the base and the original amount at `t = 0, A(0)`:

`A(t)= A(0)(1/2)^(t/t_(1/2))`

We convert `14%` to `0.14 = (A(t))/(A(0))`

`0.14=(1/2)^(t/(12.3" yr"))`

Use the natural logarithm on both sides:

`ln(0.14)=ln((1/2)^(t/(12.3" yr")))`

Use the property of logarithms that allows us to move the exponent to the outside as a coefficient:

`ln(0.14)=ln(1/2)(t/(12.3" yr"))`

Solve for t:

`t = ln(0.14)/-ln(2)12.3" yr"`

`t ~~ 34.889" yr"`