Question

how many days are required for 200. grams of radon-222 to decay to 50.0 grams? 7.64 days 1.91 days 3.82 days 11.5 days

Explanation:

Step1: Recall Radon-222 Half - Life

Radon - 222 has a half - life of approximately \(t_{1/2}=3.82\) days. The half - life formula for radioactive decay is \(N = N_0(\frac{1}{2})^{\frac{t}{t_{1/2}}}\), where \(N\) is the final amount, \(N_0\) is the initial amount, \(t\) is the time elapsed, and \(t_{1/2}\) is the half - life.

Step2: Substitute Values into the Formula

We are given that \(N_0 = 200\) grams, \(N = 50\) grams. Substitute these values into the formula:
\(50=200(\frac{1}{2})^{\frac{t}{3.82}}\)
First, divide both sides of the equation by 200:
\(\frac{50}{200}=(\frac{1}{2})^{\frac{t}{3.82}}\)
\(\frac{1}{4}=(\frac{1}{2})^{\frac{t}{3.82}}\)

Step3: Solve for \(t\)

We know that \(\frac{1}{4}=(\frac{1}{2})^2\). So, \((\frac{1}{2})^2=(\frac{1}{2})^{\frac{t}{3.82}}\)
Since the bases are the same, we can set the exponents equal to each other:
\(2=\frac{t}{3.82}\)
Multiply both sides by \(3.82\) to solve for \(t\):
\(t = 2\times3.82=7.64\)? Wait, no, wait. Wait, \(\frac{1}{4}=(\frac{1}{2})^2\), so \((\frac{1}{2})^2=(\frac{1}{2})^{\frac{t}{t_{1/2}}}\), so \(\frac{t}{t_{1/2}} = 2\), so \(t=2\times t_{1/2}\). Wait, no, wait, let's re - examine.
Wait, \(N = N_0(\frac{1}{2})^n\), where \(n\) is the number of half - lives. \(n=\frac{t}{t_{1/2}}\)
We have \(N = 50\), \(N_0 = 200\), so \(\frac{N}{N_0}=\frac{50}{200}=\frac{1}{4}\). Since \(\frac{1}{4}=(\frac{1}{2})^2\), the number of half - lives \(n = 2\).
Since each half - life of radon - 222 is \(t_{1/2}=3.82\) days? Wait, no, wait, I made a mistake earlier. Wait, radon - 222 half - life is \(3.82\) days. Wait, if we start with 200 grams and end with 50 grams, the number of half - lives:
First half - life: 200 grams \(\to\) 100 grams (after 3.82 days)
Second half - life: 100 grams \(\to\) 50 grams (after another 3.82 days)
So the total time \(t=3.82 + 3.82=7.64\) days? Wait, no, wait, let's use the formula correctly.
The formula for the number of half - lives \(n=\frac{\ln(\frac{N}{N_0})}{\ln(\frac{1}{2})}\)
\(n=\frac{\ln(\frac{50}{200})}{\ln(\frac{1}{2})}=\frac{\ln(0.25)}{\ln(0.5)}=\frac{- 1.3863}{-0.6931}=2\)
Then \(t=n\times t_{1/2}\), and \(t_{1/2}\) of radon - 222 is 3.82 days? Wait, no, I think I confused the half - life. Wait, radon - 222 has a half - life of 3.82 days. Wait, if \(n = 2\) half - lives, then \(t = 2\times3.82=7.64\) days? But wait, let's check the options.
Wait, no, wait, maybe I made a mistake in the half - life. Wait, radon - 222 has a half - life of 3.82 days. Let's re - do the calculation.
We have \(N = N_0(\frac{1}{2})^{\frac{t}{t_{1/2}}}\)
\(50 = 200(\frac{1}{2})^{\frac{t}{3.82}}\)
Divide both sides by 200: \(\frac{1}{4}=(\frac{1}{2})^{\frac{t}{3.82}}\)
Since \(\frac{1}{4}=(\frac{1}{2})^2\), then \((\frac{1}{2})^2=(\frac{1}{2})^{\frac{t}{3.82}}\), so \(\frac{t}{3.82}=2\), so \(t = 7.64\) days. Wait, but let's check the options. The first option is 7.64 days. But wait, maybe I messed up the half - life. Wait, no, radon - 222's half - life is 3.82 days. Wait, when we go from 200 to 100 (1 half - life, 3.82 days), then from 100 to 50 (second half - life, another 3.82 days), so total time is \(3.82\times2 = 7.64\) days.

Wait, but let's check the options again. The options are 7.64 days, 1.91 days, 3.82 days, 11.5 days. So the correct answer should be 7.64 days.

Answer:

7.64 days