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放射性同位体がt = 0でN0原子存在する時,時間tにおける原子数Nは次の式で表される。 N = N0 exp(-kt) 但しkは壊変定数 60Coの壊変定数がk = 0.131 /年であるとき,その半減期は何年になるか答えなさい。
When number of a certain radio isotope is N0 at t = 0, that number at t is written as the following equation.
N = N0 exp(-kt) k: radioactive decay constant Radioactive decay constant of 60Co is k = 0.131/year. Answer the half-life (year) of 60Co.