The number of nuclei of a radioactive substan | vedclass

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(B) The law of radioactive decay states that $N(t) = N_0 e^{-\lambda t}$.At $t = 0$,$N_0 = 1000$.At $t = 2 \ s$,$N(2) = 900$.So,$900 = 1000 e^{-\lambd

Vedclass · Accepted Answer

$810$

Vedclass · Answer

(B) The law of radioactive decay states that $N(t) = N_0 e^{-\lambda t}$.At $t = 0$,$N_0 = 1000$.At $t = 2 \ s$,$N(2) = 900$.So,$900 = 1000 e^{-\lambda(2)}$,which implies $e^{-2\lambda} = 0.9$.At $t = 4 \ s$,the number of nuclei $N(4) = N_0 e^{-\lambda(4)} = N_0 (e^{-2\lambda})^2$.Substituting the values: $N(4) = 1000 	imes (0.9)^2$.$N(4) = 1000 	imes 0.81 = 810$.Therefore,the number of nuclei at $t = 4 \ s$ is $810$.

The number of nuclei of a radioactive substance at time $t = 0$ is $1000$,and at time $t = 2 \ s$,it is $900$. Then,the number of nuclei at time $t = 4 \ s$ will be:

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