Answer:
A 4 half-life
B. 0.05 mg
Explanation:
A. Determination of the number of half-lives after 2.8×10⁹ years.
From the question given above,
7×10⁸ years = 1 half life
Therefore
2.8×10⁹ years = 2.8×10⁹ years × 1 half life / 7×10⁸ years
2.8×10⁹ years = 4 half life
Thus, the sample went through 4 half-lives at the end of 2.8×10⁹ years.
B. Determination of the amount of the sample remaining after 2.8×10⁹ years.
Original amount (N₀) = 0.74 mg
half life (t½) = 7×10⁸ years
Time (t) = 2.8×10⁹ years
Amount remaining (N) =?
Next, we shall determine the rate of disintegration. This can be obtained as follow:
half life (t½) = 7×10⁸ years
Decay constant (K) =?
K = 0.693 / t½
K = 0.693 / 7×10⁸
K = 9.9×10¯¹⁰ /year
Finally, we shall determine the amount remaining as follow:
Original amount (N₀) = 0.74 mg
Time (t) = 2.8×10⁹ years
Decay constant (K) = 9.9×10¯¹⁰ /year
Amount remaining (N) =?
Log (N₀/N) = kt / 2.303
Log (0.74/N) = 9.9×10¯¹⁰×2.8×10⁹ /2.303
Log (0.74/N) = 2.772 / 2.303
Log (0.74/N) = 1.2036
Take the antilog of 1.2036
0.74/N = antilog (1.2036)
0.74 / N = 15.98
Cross multiply
0.74 = N × 15.98
Divide both side by 15.98
N = 0.74 / 15.98
N = 0.05 mg
Thus, 0.05 mg of the sample will remain after 2.8×10⁹ years
