Answer:
3.57 mg
Step-by-step explanation:
The following data were obtained from the question:
Half life (t½) = 12.3 years
Original amount (N₀) = 57.0 mg
Time (t) = 49.2 years
Amount remaining (N) =?
Next, we shall determine the rate of disintegration. This can be obtained as follow:
Half life (t½) = 12.3 years
Decay constant (K) =.?
K = 0.693 / t½
K = 0.693 / 12.3
K = 0.0563 /year
Finally, we shall determine the amount remaining after 49.2 years as shown below:
Original amount (N₀) = 57.0 mg
Time (t) = 49.2 years
Decay constant (K) = 0.0563 /year
Amount remaining (N) =?
Log (N₀/N) = kt / 2.303
Log (57/N) = (0.0563 × 49.2) / 2.303
Log (57/N) = 2.76996 / 2.303
Log (57/N) = 1.2028
Take the anti log of 1.2028
(57/N) = anti log (1.2028)
57/N = 15.95
Cross multiply
57 = N × 15.95
Divide both side by 15.95
N = 57 / 15.95
N = 3.57 mg
Thus, 3.57 mg of the nuclide will be remaining after 49.2 years.