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The mean lives of a radioactive substance are 1620 year and 405 year for alpha-emission and beta-emission respectively. Find the time during which three - forth of a sample will decay if it is decayed both by alpha -emission and beta-emission simultaneously. |
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Answer» Solution :The DECAY constant `lambda` is the reciprocal of the mean life, `tau` Thus, `lambda_alpha=1/1620` PER year and `lambda_beta=1/405` Per year `therefore` Total decay constant , `lambda=lambda_alpha+lambda_beta` `lambda=1/1620+1/405=1/324` year We know that `N=N_0 e^(-lambdat)` When `3/4`th part of the sample has disintegrated, `N=N_0//4 therefore N_0/4 = N_0e^(-lambdat)` or `e^(lambdat)`=4 Taking logarithm on both SIDES , we get `lambdat=log_e 4` `t=1/lambdalog_e 2^2 = 2/lambda log_e2` =2 x 324 x 0.693 = 449 years. |
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