Question

a sample contains 16.75 g of the radioisotope u - 236 and 50.25 g of its daughter isotope, th - 232. how long did it take for decay to take place if one half - life of u - 236 is 23 million years? 115 million years 92 million years 69 million years 46 million years

Explanation:

Step1: Calculate the total mass of the original isotope

The total mass of the original isotope (before decay) is the sum of the remaining radioisotope (U - 236) and the daughter isotope (Th - 232). So, \(m_{original}=16.75 + 50.25=67\space g\)

Step2: Determine the fraction of the remaining radioisotope

The fraction of the remaining radioisotope (\(N/N_0\)) is given by \(\frac{N}{N_0}=\frac{16.75}{67}=\frac{1}{4}\)

Step3: Relate the fraction to the number of half - lives

We know that \(N = N_0\times(\frac{1}{2})^n\), where \(n\) is the number of half - lives. If \(\frac{N}{N_0}=\frac{1}{4}=(\frac{1}{2})^2\), then \(n = 2\)

Step4: Calculate the time of decay

Since the half - life \(t_{1/2}=23\space million\space years\) and \(t=n\times t_{1/2}\), substituting \(n = 2\) and \(t_{1/2}=23\space million\space years\), we get \(t=2\times23 = 46\space million\space years\)

Answer:

46 million years