Question

if you start with 120 - gram sample of radium. what percent will be left after 44 days?

Explanation:

1. Explanation:

• Step 1: Recall the radioactive - decay formula
• The radioactive - decay formula is \(N = N_0(\frac{1}{2})^{\frac{t}{T_{1/2}}}\), where \(N\) is the amount of the substance remaining after time \(t\), \(N_0\) is the initial amount of the substance, \(t\) is the time elapsed, and \(T_{1/2}\) is the half - life of the substance. The half - life of Radium - 226 is \(T_{1/2}=1600\) years. First, convert the time \(t = 44\) days to years. Since there are 365 days in a year, \(t=\frac{44}{365}\) years.
• \(t=\frac{44}{365}\approx0.12\) years.
• Step 2: Calculate the fraction of the remaining substance
• Substitute \(N_0\) (initial amount), \(t\) (time elapsed), and \(T_{1/2}\) (half - life) into the formula \(N = N_0(\frac{1}{2})^{\frac{t}{T_{1/2}}}\).
• \(\frac{N}{N_0}=(\frac{1}{2})^{\frac{t}{T_{1/2}}}\), substituting \(t = 0.12\) years and \(T_{1/2}=1600\) years, we get \(\frac{N}{N_0}=(\frac{1}{2})^{\frac{0.12}{1600}}\).
• Let \(x=\frac{0.12}{1600}=7.5\times10^{-5}\). Then \(\frac{N}{N_0}=2^{-x}\). Using the property \(a^{-x}=\frac{1}{a^{x}}\), and the approximation \(2^{-x}\approx1 - x\ln(2)\) (for \(x\ll1\)).
• \(\ln(2)\approx0.693\), so \(\frac{N}{N_0}\approx1-(7.5\times10^{-5})\times0.693\).
• \(\frac{N}{N_0}\approx1 - 5.2\times10^{-5}\).
• To convert this to a percentage, multiply by 100. The percentage of the remaining substance is \(P=\frac{N}{N_0}\times100\approx(1 - 5.2\times10^{-5})\times100 = 99.9948\%\).

1. Answer:

• Approximately \(99.99\%\)

Answer:

1. Explanation:

• Step 1: Recall the radioactive - decay formula
• The radioactive - decay formula is \(N = N_0(\frac{1}{2})^{\frac{t}{T_{1/2}}}\), where \(N\) is the amount of the substance remaining after time \(t\), \(N_0\) is the initial amount of the substance, \(t\) is the time elapsed, and \(T_{1/2}\) is the half - life of the substance. The half - life of Radium - 226 is \(T_{1/2}=1600\) years. First, convert the time \(t = 44\) days to years. Since there are 365 days in a year, \(t=\frac{44}{365}\) years.
• \(t=\frac{44}{365}\approx0.12\) years.
• Step 2: Calculate the fraction of the remaining substance
• Substitute \(N_0\) (initial amount), \(t\) (time elapsed), and \(T_{1/2}\) (half - life) into the formula \(N = N_0(\frac{1}{2})^{\frac{t}{T_{1/2}}}\).
• \(\frac{N}{N_0}=(\frac{1}{2})^{\frac{t}{T_{1/2}}}\), substituting \(t = 0.12\) years and \(T_{1/2}=1600\) years, we get \(\frac{N}{N_0}=(\frac{1}{2})^{\frac{0.12}{1600}}\).
• Let \(x=\frac{0.12}{1600}=7.5\times10^{-5}\). Then \(\frac{N}{N_0}=2^{-x}\). Using the property \(a^{-x}=\frac{1}{a^{x}}\), and the approximation \(2^{-x}\approx1 - x\ln(2)\) (for \(x\ll1\)).
• \(\ln(2)\approx0.693\), so \(\frac{N}{N_0}\approx1-(7.5\times10^{-5})\times0.693\).
• \(\frac{N}{N_0}\approx1 - 5.2\times10^{-5}\).
• To convert this to a percentage, multiply by 100. The percentage of the remaining substance is \(P=\frac{N}{N_0}\times100\approx(1 - 5.2\times10^{-5})\times100 = 99.9948\%\).

1. Answer:

• Approximately \(99.99\%\)