To solve this problem, we can use the formula for exponential decay, which models the reduction in quantity of a radioactive substance over time. The decay formula is given by:
[tex]\[ N(t) = N_0 \cdot e^{-\lambda t} \][/tex]
where:
- [tex]\( N(t) \)[/tex] is the remaining amount after time [tex]\( t \)[/tex].
- [tex]\( N_0 \)[/tex] is the original amount of the substance.
- [tex]\( e \)[/tex] is the base of the natural logarithm.
- [tex]\( \lambda \)[/tex] is the decay constant.
- [tex]\( t \)[/tex] is the time.
Given data:
- Decay constant, [tex]\( \lambda = 0.046 \)[/tex] per hour.
- Remaining amount after time [tex]\( t \)[/tex], [tex]\( N(t) = 50 \)[/tex] mg.
- Time elapsed, [tex]\( t = 3 \)[/tex] hours.
We need to find the original amount, [tex]\( N_0 \)[/tex]. Rearranging the decay formula to solve for [tex]\( N_0 \)[/tex] gives:
[tex]\[ N_0 = \frac{N(t)}{e^{-\lambda t}} \][/tex]
First, calculate the term [tex]\( e^{-\lambda t} \)[/tex]:
[tex]\[ e^{-\lambda t} = e^{-0.046 \cdot 3} \][/tex]
By evaluating the exponent,
[tex]\[ e^{-0.046 \cdot 3} \approx 0.8711 \][/tex]
Next, we use this value to find [tex]\( N_0 \)[/tex]:
[tex]\[ N_0 = \frac{50 \; \text{mg}}{0.8711} \][/tex]
Dividing the remaining amount by the calculated exponential term:
[tex]\[ N_0 \approx \frac{50}{0.8711} \approx 57.40 \; \text{mg} \][/tex]
Therefore, the original amount of the radioactive isotope present was approximately [tex]\( 57.40 \; \text{mg} \)[/tex].
