write and solve an algebraic equation to find the original mass of the sample of the radioactive isotope uranium -235, which has a half-life of 704 million years your sample is about 700 million years old and weighs 120 grams
To solve for the original mass of the sample of uranium-235, we can use the formula for exponential decay:
\[A = A_0 \times \left(\frac{1}{2}\right)^{\frac{t}{h}}\]
Where:
A = Final mass of the sample (120 grams in this case)
A₀ = Original mass of the sample (what we want to find)
t = Time elapsed (700 million years in this case)
h = Half-life of the radioactive isotope (704 million years in this case)
Plugging in the given values into the equation, we can solve for A₀.
\[120 = A_0 \times \left(\frac{1}{2}\right)^{\frac{700}{704}}\]
To solve for A₀, we can isolate it by dividing both sides of the equation by \(\left(\frac{1}{2}\right)^{\frac{700}{704}}\).
\[A_0 = \frac{120}{\left(\frac{1}{2}\right)^{\frac{700}{704}}}\]
Simplifying the fraction:
\[A_0 = 120 \times \left(\frac{2}{1}\right)^{\frac{700}{704}}\]
Now we need to evaluate the value of \(\left(\frac{2}{1}\right)^{\frac{700}{704}}\).
\(\left(\frac{2}{1}\right)^{\frac{700}{704}}\) can be simplified as follows:
\(\left(\frac{2}{1}\right)^{\frac{700}{704}} = \left(\frac{2}{1}\right)^{1 - \frac{4}{704}} = \left(\frac{2}{1}\right)^{1 - \frac{1}{176}} = \left(\frac{2}{1}\right)^{\frac{175}{176}}\)
Since we know that \(\left(\frac{a}{b}\right)^{\frac{c}{d}} = \left(\frac{a^c}{b^d}\right)\), we have:
\(\left(\frac{2}{1}\right)^{\frac{175}{176}} = \frac{2^{175}}{1^{176}} = 2^{175}\)
Now we can substitute this value back into the equation to find A₀:
\[A_0 = 120 \times 2^{175}\]
Calculating this value will give us the original mass of the uranium-235 sample.