If 20 gm of a radioactive substance due to radioactive decay reduces to 10 gm in 4 minutes, then in what time 80 gm of the same substance will reduce to 10 gm-

Options

A. 8 minutes
B. 12 minutes
C. 16 minutes
D. 20 minutes

Answer

To solve the problem step by step, we will use the concept of half-life in radioactive decay.

### Step 1: Determine the half-life of the substance
We know that 20 grams of the radioactive substance reduces to 10 grams in 4 minutes. This indicates that the substance has undergone one half-life.

- **Half-life (t_half)** = 4 minutes

### Step 2: Understand the relationship between initial mass, final mass, and half-life
The formula for radioactive decay can be expressed as:
\[ m = m_0 \left(\frac{1}{2}\right)^{\frac{T}{t_{half}}} \]
Where:
- \( m \) = final mass
- \( m_0 \) = initial mass
- \( T \) = time elapsed
- \( t_{half} \) = half-life

### Step 3: Set up the equation for the second scenario
Now, we need to find out how long it will take for 80 grams of the substance to reduce to 10 grams.

- Initial mass \( m_0 = 80 \) grams
- Final mass \( m = 10 \) grams

Using the decay formula:
\[ 10 = 80 \left(\frac{1}{2}\right)^{\frac{T}{4}} \]

### Step 4: Simplify the equation
First, divide both sides by 80:
\[ \frac{10}{80} = \left(\frac{1}{2}\right)^{\frac{T}{4}} \]
This simplifies to:
\[ \frac{1}{8} = \left(\frac{1}{2}\right)^{\frac{T}{4}} \]

### Step 5: Express \(\frac{1}{8}\) in terms of powers of \(\frac{1}{2}\)
We know that:
\[ \frac{1}{8} = \left(\frac{1}{2}\right)^{3} \]

### Step 6: Set the exponents equal to each other
Now we can equate the exponents:
\[ \frac{T}{4} = 3 \]

### Step 7: Solve for \( T \)
Multiply both sides by 4 to solve for \( T \):
\[ T = 3 \times 4 = 12 \text{ minutes} \]

### Conclusion
The time it takes for 80 grams of the substance to reduce to 10 grams is **12 minutes**.

---