After 2 h has elapsed, one-sixteenth of the original quantity of a certain radionuclide remains undecayed. The half-life of this radionuclide is

Question

Quizplus · Accepted Answer

If one-sixteenth of the original quantity remains undecayed after 2 hours, then 15/16 of the original quantity has decayed.

Let N0 be the original quantity and N be the quantity remaining after 2 hours. Then we have:

N/N0 = 1/16

N0 - N = (15/16)N0

Solving these equations, we get:

N0/16 = N

N0 - N = (15/16)N0

Therefore, (1/16)N0 = (15/16)N0

This implies that the fraction of the original quantity that has decayed is 15/16.

Now, we use the formula for the half-life of a radionuclide:

N/N0 = (1/2)^(t/T)

where t is the time elapsed and T is the half-life.

Since 15/16 of the original quantity has decayed, we have:

N/N0 = 1/16 = (1/2)^(t/T)

Taking the logarithm of both sides and solving for T, we get:

T = t/log2(16)

T = t/4

Since t = 2 hours = 120 minutes, we have:

T = 120/4 = 30 minutes

Therefore, the half-life of the radionuclide is 30 minutes. The best choice is B.

After 2 H Has Elapsed, One-Sixteenth of the Original Quantity