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A radioactive substance has 6.0 xx 10^(18)active nuclei initially . What time required for the active nuclei of the same substance to become 1.0 xx 10^(18)if it.s half-life is 40 s. |
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Answer» Solution :The NUMBER of ACTIVE nuclei at any instant of time t , `N_0/N= e^(lamdat) "" log_(e)(N_0/N)= lamdat` `:.t= (log_e((N_0)/(N)))/lamda=(2.303 log_(10)(N_0/N))/lamda` In this problem , the INITIAL number of active nuclei, `N_(0) = 6.0 xx10^(18), N=1.0 xx10^(18), T = 40s` `lamda=(0.693)/T=(0.693)/40=1.733xx10^(-2)s^(-1)`. `t= (2.303log_(10)((6.0xx10^(18))/(1.0xx10^(18))))/(1.733xx10^(-2))` `= (2.303log_(10)(6))/(1.733xx10^(-2))=(2.303xx0.7782)/(1.733xx10^(-2))=103.4s.` |
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