Question

一道半衰期的数学题，求解！不胜感激！！

In the early 1960s, radioactive strontium-90 was released during atmospheric testing of nuclear weapons and got into the bones of people alive at the time. If the half-life of strontium-90 is 29 years, what fraction of the strontium-90 absorbed in 1962 remained in people's bones in 2003?

可以用分数或者小数表示，需要过程，谢谢！

Answer 1

N/N0=e^(-λt)
=e^(-(0.693/T)t)
=e^(-(0.693/29)(2003-1962))
=e^(-0.979759)
=0.3754
=37.54%