An isotope of radon gas, Rn-222, undergoes decay with a half-life of 3.8 days. How long does it take for the amount of Rn-222 to be decreased to one-sixteenth of the original amount?
The Correct Answer and Explanation is:
To determine how long it takes for the amount of Rn-222 to decrease to one-sixteenth of the original amount, we can use the concept of half-life and apply it to the decay process.
Step 1: Understanding the concept of half-life
The half-life of an isotope is the time it takes for half of the substance to decay. In this case, we are given that the half-life of Rn-222 is 3.8 days.
Step 2: Using the decay formula
The general decay formula is:N(t)=N0(12)tT1/2N(t) = N_0 \left(\frac{1}{2}\right)^{\frac{t}{T_{1/2}}}N(t)=N0(21)T1/2t
Where:
- N(t)N(t)N(t) is the remaining amount of substance after time ttt.
- N0N_0N0 is the initial amount of substance.
- T1/2T_{1/2}T1/2 is the half-life (3.8 days in this case).
- ttt is the time elapsed.
We want to find the time ttt when the remaining amount of radon is one-sixteenth of the original amount, meaning N(t)=116N0N(t) = \frac{1}{16} N_0N(t)=161N0.
Step 3: Setting up the equation
Substitute N(t)=116N0N(t) = \frac{1}{16} N_0N(t)=161N0 into the decay equation:116N0=N0(12)t3.8\frac{1}{16} N_0 = N_0 \left(\frac{1}{2}\right)^{\frac{t}{3.8}}161N0=N0(21)3.8t
Step 4: Simplifying the equation
Cancel N0N_0N0 from both sides of the equation (assuming N0≠0N_0 \neq 0N0=0):116=(12)t3.8\frac{1}{16} = \left(\frac{1}{2}\right)^{\frac{t}{3.8}}161=(21)3.8t
Now express 116\frac{1}{16}161 as (12)4\left(\frac{1}{2}\right)^4(21)4:(12)4=(12)t3.8\left(\frac{1}{2}\right)^4 = \left(\frac{1}{2}\right)^{\frac{t}{3.8}}(21)4=(21)3.8t
Step 5: Solving for ttt
Since the bases are the same, we can set the exponents equal to each other:4=t3.84 = \frac{t}{3.8}4=3.8t
Now solve for ttt:t=4×3.8=15.2 dayst = 4 \times 3.8 = 15.2 \text{ days}t=4×3.8=15.2 days
Conclusion:
It will take 15.2 days for the amount of Rn-222 to decrease to one-sixteenth of the original amount.
