`A` and `B` are two radioactive substances whose half lives are `1` and `2` years respectively. Initially `10 gm` of `A` and `1 gm` of `B` is taken. The time (approximate) after which they will have same quantity remaining is.

Options

A. 6.62 years
B. 5 years
C. 3.2 years
D. 7 years

Answer

To solve the problem, we need to find the time at which the remaining quantities of two radioactive substances, A and B, become equal.

### Step-by-Step Solution:

1. **Understand the Half-Life Concept**:
- The half-life of a substance is the time required for half of the substance to decay.
- For substance A, the half-life (T₁/₂) is 1 year.
- For substance B, the half-life (T₁/₂) is 2 years.

2. **Initial Amounts**:
- Initially, we have 10 grams of substance A (N₀(A) = 10 g).
- Initially, we have 1 gram of substance B (N₀(B) = 1 g).

3. **Radioactive Decay Formula**:
- The remaining quantity of a radioactive substance after time t can be calculated using the formula:
\[ N(t) = N_0 \left(\frac{1}{2}\right)^{\frac{t}{T_{1/2}}} \]
- For substance A:
\[ N_A(t) = 10 \left(\frac{1}{2}\right)^{\frac{t}{1}} = 10 \left(\frac{1}{2}\right)^{t} \]
- For substance B:
\[ N_B(t) = 1 \left(\frac{1}{2}\right)^{\frac{t}{2}} = 1 \left(\frac{1}{2}\right)^{\frac{t}{2}} \]

4. **Set the Remaining Quantities Equal**:
- We want to find the time t when \( N_A(t) = N_B(t) \):
\[ 10 \left(\frac{1}{2}\right)^{t} = 1 \left(\frac{1}{2}\right)^{\frac{t}{2}} \]

5. **Simplify the Equation**:
- Divide both sides by \( \left(\frac{1}{2}\right)^{\frac{t}{2}} \):
\[ 10 \left(\frac{1}{2}\right)^{t - \frac{t}{2}} = 1 \]
- This simplifies to:
\[ 10 \left(\frac{1}{2}\right)^{\frac{t}{2}} = 1 \]

6. **Isolate the Exponential Term**:
- Rearranging gives:
\[ \left(\frac{1}{2}\right)^{\frac{t}{2}} = \frac{1}{10} \]

7. **Take Logarithm**:
- Taking logarithm on both sides:
\[ \frac{t}{2} \log\left(\frac{1}{2}\right) = \log\left(\frac{1}{10}\right) \]
- Since \( \log\left(\frac{1}{2}\right) = -\log(2) \):
\[ -\frac{t}{2} \log(2) = -\log(10) \]

8. **Solve for t**:
- Rearranging gives:
\[ \frac{t}{2} = \frac{\log(10)}{\log(2)} \]
- Therefore:
\[ t = 2 \cdot \frac{\log(10)}{\log(2)} \]

9. **Calculate the Value**:
- Using \( \log(10) \approx 1 \) and \( \log(2) \approx 0.3010 \):
\[ t \approx 2 \cdot \frac{1}{0.3010} \approx 6.64 \text{ years} \]

### Final Answer:
The time after which both substances will have the same quantity remaining is approximately **6.64 years**.