Answer:
6.23 g.
Step-by-step explanation:
From the question given above, the following data were obtained:
Original amount (N₀) = 100 g
Time (t) = 240 days
Half life (t½) = 60 days
Amount remaining (N) =?
Next, we shall determine the decay constant (K). This can be obtained as follow:
Half life (t½) = 60 days
Decay constant (K) =.?
Decay constant (K) = 0.693/Half life (t½)
K = 0.693/60
K = 0.01155 /day
Finally, we shall determine the amount of gamma emitter l-125 that remains after 240 days as follow:
Original amount (N₀) = 100 g
Time (t) = 240 days
Half life (t½) = 60 days
Decay (K) = 0.01155 /day
Amount remaining (N) =?
Log (N₀/N) = kt/2.3
Log (100/N) = (0.01155 × 240)/2.3
Log (100/N) = 2.772/2.3
Log (100/N) = 1.2052
Take the anti log of 1.2052
(100/N) = antilog (1.2052)
100/N= 16.04
Cross multiply
100 = N × 16.04
Divide both side by 16.04
N = 100/16.04
N = 6.23 g
Therefore, the amount of the gamma emitter l-125 that remains after 240 days is 6.23 g