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Question
the isotope titanium - 44 decays into scandium - 44, with a half - life of 60 years. in a sample of titanium - 44 containing 2×10¹⁰ atoms, how many titanium - 44 and scandium - 44 atoms would there be in the sample after 240 years?
step 1: how many half - lives will occur? 1 half - life = ______ years
240 years ÷ ____ years/half - life = __ half - lives
step 2: how many titanium - 44 atoms remain?
step 3: how many scandium - 44 atoms have been produced?
2×10¹⁰−1.25×10⁹ = ______×10¹⁰ atoms of scandium - 44
Step1: Calculate the number of half - lives
The half - life of titanium - 44 is 60 years.
To find the number of half - lives in 240 years, we use the formula \(n=\frac{t}{T}\), where \(t = 240\) years (total time) and \(T=60\) years (half - life).
\(n=\frac{240}{60}=4\) half - lives.
Step2: Calculate the remaining titanium - 44 atoms
The formula for the number of remaining atoms \(N = N_0\times(\frac{1}{2})^n\), where \(N_0 = 2\times10^{10}\) (initial number of atoms) and \(n = 4\) (number of half - lives).
We can also fill the table:
- When \(n = 0\), \(N=2\times10^{10}\) (so \(2\times10^{10}\), \(10\) has exponent \(10\))
- When \(n = 1\), \(N = 2\times10^{10}\times\frac{1}{2}=1\times10^{10}\) (so \(1\times10^{10}\))
- When \(n = 2\), \(N=2\times10^{10}\times(\frac{1}{2})^2 = 0.5\times10^{10}=5\times10^{9}\)
- When \(n = 3\), \(N=2\times10^{10}\times(\frac{1}{2})^3=0.25\times 10^{10}=2.5\times10^{9}\)
- When \(n = 4\), \(N = 2\times10^{10}\times(\frac{1}{2})^4=1.25\times10^{9}\)
Step3: Calculate the number of scandium - 44 atoms
The number of scandium - 44 atoms \(N_{Sc}=N_0 - N\)
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- Step 1: \(1\) half - life \(= 60\) years; \(240\) years \(\div60\) years/half - life \( = 4\) half - lives
- Step 2: The table:
| Number of Half - lives | Number of Titanium - 44 Atoms Remaining |
|---|---|
| \(1\) | \(1\times10^{10}\) |
| \(2\) | \(5\times10^{9}\) |
| \(3\) | \(2.5\times10^{9}\) |
| \(4\) | \(1.25\times10^{9}\) |
- Step 3: \(2\times10^{10}-1.25\times10^{9}=1.875\times10^{10}\) atoms of scandium - 44.