The number of a certain radioactive nuclide present in a sample decays from2.0 × 10² to 6.0 × 10¹ in 23 minutes. What is the half-life of this radioactive species? A) 1.6 × 10¹ minutes B) 5.2 × 10^-2 minutes C) 1.3 × 10¹ minutes D) 5.3 × 10¹ minutes E) 3.0 minutes

Question

Quizplus · Accepted Answer

We can use the following formula for radioactive decay:

N(t) = N0*(1/2)^(t/T1/2)

where N(t) is the amount of the radioactive nuclide at a given time t, N0 is the initial amount of the nuclide, T1/2 is the half-life of the nuclide.

We can plug in the given values:

N0 = 2.0 × 10^12
N(t) = 6.0 × 10^11
t = 23 minutes

We can solve for T1/2:

N(t)/N0 = (1/2)^(t/T1/2)
6.0 × 10^11 / 2.0 × 10^12 = (1/2)^(23/T1/2)
0.3 = (1/2)^(23/T1/2)

Take the logarithm of both sides:

log(0.3) = log[(1/2)^(23/T1/2)]
log(0.3) = (23/T1/2) * log(1/2)
log(0.3) = -(23/T1/2) * log(2)
-(log(0.3)/log(2)) = 23/T1/2
T1/2 = -23 / (log(0.3)/log(2))

T1/2 = 1.3 × 10^11 seconds or 1.3 × 10^11 / 60 = 2.17 × 10^9 minutes or 1.3 × 10^11 / 3600 = 3.6 × 10^7 hours

The closest answer choice in minutes is C) 1.3 × 10^11 / 60 = 2.17 × 10^9 minutes, which rounds to 1.3 × 10^11 minutes. Therefore, the answer is C.

The Number of a Certain Radioactive Nuclide Present in a Sample