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The number of disintegration per minute of a certain radioactive substance are 6050 initially and 4465 at the end of one hour. Calculate the decay constant and the half life of the substance. |
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Answer» We know that R = R0 e-λt 6080 = R0 eλt 4465 = R0 e-λ(t+60) \(\frac {6080}{4465} = \frac {e^{-λt}}{e^{-λt}} e^{-60λ}\) e+60λ =\(\frac {6080}{4465} \) e+60λ= 1.354 60λ = ln (1.354) λ = \(\frac {1}{60}\) ln (1.354) λ = \(\frac {0.30}{60}\) Decay constant λ = 0.005 m Half life T1/2 = \(\frac {0.693}{λ}\) T1/2 = \(\frac {0.693}{0.005}\) T1/2 = 138.6 min |
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