Question

1. how much of a 100.0 g sample of ¹⁹⁸au is left after 8.10 days if its half - life is 2.70 days?
2. a 50.0 g sample of ¹⁶n decays to 12.5 g in 14.4 seconds. what is its half - life?
3. the half - life of ⁴²k is 12.4 hours. how much of a 750 g sample is left after 62.0 hours?
4. what is the half - life of ⁹⁹tc if a 500 g sample decays to 62.5 g in 639,000 years?
5. the half - life of ²³²th is 1.4 x 10¹⁰ years. if there are 25.0 g of the sample left after 2.8 x 10¹⁰ years, how many grams were in the original sample?

Explanation:

Step1: Calculate the number of half - lives

The formula for the number of half - lives \(n=\frac{t}{t_{1/2}}\), where \(t\) is the time elapsed and \(t_{1/2}\) is the half - life.
For problem 1: \(n=\frac{8.10}{2.70}=3\)
For problem 2: Let the number of half - lives be \(n\). We know that \(N = N_0\times(\frac{1}{2})^n\), where \(N_0 = 50.0\space g\), \(N=12.5\space g\). So \(12.5=50.0\times(\frac{1}{2})^n\), \(\frac{12.5}{50.0}=(\frac{1}{2})^n\), \(\frac{1}{4}=(\frac{1}{2})^n\), \(n = 2\). Then \(t_{1/2}=\frac{t}{n}=\frac{14.4}{2}=7.2\space s\)
For problem 3: \(n=\frac{62.0}{12.4}=5\)
For problem 4: \(N = N_0\times(\frac{1}{2})^n\), \(62.5=500\times(\frac{1}{2})^n\), \(\frac{62.5}{500}=(\frac{1}{2})^n\), \(\frac{1}{8}=(\frac{1}{2})^n\), \(n = 3\). Then \(t_{1/2}=\frac{t}{n}=\frac{639000}{3}=213000\space years\)
For problem 5: \(n=\frac{2.8\times 10^{10}}{1.4\times 10^{10}}=2\). Using \(N = N_0\times(\frac{1}{2})^n\), \(25.0=N_0\times(\frac{1}{2})^2\), \(N_0=25.0\times4 = 100\space g\)

Step2: Calculate the remaining amount or half - life

For problem 1: Using \(N = N_0\times(\frac{1}{2})^n\), where \(N_0 = 100.0\space g\), \(n = 3\). So \(N=100.0\times(\frac{1}{2})^3=100.0\times\frac{1}{8}=12.5\space g\)
For problem 3: Using \(N = N_0\times(\frac{1}{2})^n\), where \(N_0 = 750\space g\), \(n = 5\). So \(N=750\times(\frac{1}{2})^5=750\times\frac{1}{32}=23.4375\space g\)

Answer:

1. \(12.5\space g\)
2. \(7.2\space s\)
3. \(23.4375\space g\)
4. \(213000\space years\)
5. \(100\space g\)