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In a radioactive material the activity at time $t_1$ is $R_1$ and at a later time $t_2$, it is $R_2$. If the decay constant of the material is $\lambda$, then
A
$R_1 = R_2\,e^{-\lambda(t_1 - t_2)}$
B
$R_1 = R_2\,e^{\lambda(t_1 - t_2)}$
C
$R_1 = R_2\left(\dfrac{t_2}{t_1}\right)$
D
$R_1 = R_2$
Detailed Solution
The activity of a radioactive sample decreases exponentially with time: $R = R_0 e^{-\lambda t}$
At time $t_1$: $R_1 = R_0 e^{-\lambda t_1}$
At time $t_2$: $R_2 = R_0 e^{-\lambda t_2}$
Dividing: $\dfrac{R_1}{R_2} = \dfrac{e^{-\lambda t_1}}{e^{-\lambda t_2}} = e^{-\lambda(t_1 - t_2)}$
$\Rightarrow R_1 = R_2\,e^{-\lambda(t_1 - t_2)}$
Since $t_2 \gt t_1$, the exponent is positive and $R_1 \gt R_2$, as expected.
At time $t_1$: $R_1 = R_0 e^{-\lambda t_1}$
At time $t_2$: $R_2 = R_0 e^{-\lambda t_2}$
Dividing: $\dfrac{R_1}{R_2} = \dfrac{e^{-\lambda t_1}}{e^{-\lambda t_2}} = e^{-\lambda(t_1 - t_2)}$
$\Rightarrow R_1 = R_2\,e^{-\lambda(t_1 - t_2)}$
Since $t_2 \gt t_1$, the exponent is positive and $R_1 \gt R_2$, as expected.