If the rate of decay is 1% per year, the number of grams present after 12 years is found from:
\(52\times(0.99)^{12}=46\ grams\)
An alternative solution can use the following equation:
\(N(t)= N_{0} e^{-\lambda t}\)
Plugging in given values, we get:
\(N=52\times e^{-0.01\times t}=46\ grams\)
