A 280, text{day} old radioactive substance sh | vedclass

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(B) The activity of a radioactive substance at time $t$ is given by $A = A_0 e^{-\lambda t}$, where $A_0$ is the initial activity.Let $A_1 = 6000\,

Vedclass · Accepted Answer

$24000$

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(B) The activity of a radioactive substance at time $t$ is given by $A = A_0 e^{-\lambda t}$, where $A_0$ is the initial activity.Let $A_1 = 6000\, 	ext{dps}$ at $t_1 = 280\, 	ext{days}$.Let $A_2 = 3000\, 	ext{dps}$ at $t_2 = 280 + 140 = 420\, 	ext{days}$.Since the activity drops from $6000\, 	ext{dps}$ to $3000\, 	ext{dps}$ in $140\, 	ext{days}$, the half-life $T_{1/2}$ is $140\, 	ext{days}$.Using the relation $A = A_0 (1/2)^{t/T_{1/2}}$, we find the initial activity $A_0$ at $t = 0$ using the state at $t_1 = 280\, 	ext{days}$:$6000 = A_0 (1/2)^{280/140}$$6000 = A_0 (1/2)^2$$6000 = A_0 / 4$$A_0 = 6000 	imes 4 = 24000\, 	ext{dps}$.

$A$ $280\, \text{day}$ old radioactive substance shows an activity of $6000\, \text{dps}$. $140\, \text{days}$ later, its activity becomes $3000\, \text{dps}$. What was its initial activity? ......... $\text{dps}$

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