99% quantity of a radioactive substance decays between

Options

A. 8 and 9 half lives
B. 7 and 8 half lives
C. 9 and 10 half lives
D. 6 and 7 half lives

Explanation

To solve the problem of determining between which half-lives 99% of a radioactive substance decays, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Problem**: We need to find out how many half-lives it takes for 99% of a radioactive substance to decay. If 99% decays, then only 1% remains.

2. **Setting Up the Equation**: Let \( N_0 \) be the initial quantity of the substance. After \( n \) half-lives, the remaining quantity \( N \) can be expressed as:
\[ N = N_0 \left(\frac{1}{2}\right)^n \]
Since 99% has decayed, we have:
\[ N = N_0 - 0.99 N_0 = 0.01 N_0 \]

3. **Equating the Two Expressions**: We can set the two expressions for \( N \) equal to each other:
\[ N_0 \left(\frac{1}{2}\right)^n = 0.01 N_0 \]
Dividing both sides by \( N_0 \) (assuming \( N_0 \neq 0 \)):
\[ \left(\frac{1}{2}\right)^n = 0.01 \]

4. **Rewriting the Right Side**: We can express 0.01 as a power of 2:
\[ 0.01 = \frac{1}{100} = \frac{1}{10^2} = 10^{-2} \]
However, we will convert it to a base of 2:
\[ 0.01 = \frac{1}{2^n} \implies 2^n = 100 \]

5. **Finding \( n \)**: To find \( n \), we can take the logarithm (base 2) of both sides:
\[ n = \log_2(100) \]
We know that \( 2^6 = 64 \) and \( 2^7 = 128 \). Therefore, \( n \) must be between 6 and 7.

6. **Conclusion**: Since \( n \) is between 6 and 7, we conclude that 99% of the radioactive substance decays between the 6th and 7th half-lives.

### Final Answer:
99% of the quantity of a radioactive substance decays between the 6th and 7th half-lives.