Question

All living things are radioactive due to the presence of 14 6 C which undergoes beta decay, emitting an electron. The half-life of 14 6 C is 5730 y. In live specimens, the level of 14 6 C present is constant. After death, the decay of 14 6 C slowly reduces the amount of 14 6 C present. Each gram of a live specimen emits 15.3 electrons per minute. An animal bone found at an archaeological site emits 86 electrons per minute. Its mass is 13.2 g. How old (years) is the bone? (How long ago did the animal die?) Give the answer to 2 significant figures.

Answer 1

half-life of 14 6 C is 5730 y

In live specimens, the level of 14 6 C present is constant.

Each gram of a live specimen emits 15.3 electrons per minute

A radioactive atom can decay by emitting a beta particle which is a fast moving electron

animal bone emits 86 electrons per minute

we know decay equation is of the form

derivating w r t time

this give the decay rate

at t =0 rate = 15.3 electon per minute per gram

at other time t rate = 86/13.2 = 6.515 electrons per minute per gram

putting value

we get ===> 6.51 = 15.3   

   = 0.693/5730 = 0.000121

we get 0.4255 = e^{-0.000121 * t}

   = -0.8545 = -0.000121 * t ====> t = 0.8545 / 0.000121 = 7062 years