If the half lives of a radioactive element for `alpha` and `beta` decay are 4 year and 12 years respectively, then the percentage of the element that remains after 12 year will be

Options

A. (a)`6.25%`
B. (b)`12.5%`
C. (c)`25%`
D. (d)`50%`

Explanation

To solve the problem, we need to determine the remaining percentage of a radioactive element after 12 years, given its half-lives for alpha and beta decay.

### Step 1: Understand the half-life concept
The half-life of a radioactive element is the time taken for half of the radioactive substance to decay. After each half-life, the amount of the substance reduces to half of its previous amount.

### Step 2: Calculate the decay constant (λ) for both decays
The decay constant (λ) can be calculated using the formula:
\[ \lambda = \frac{\ln(2)}{T_{1/2}} \]
where \( T_{1/2} \) is the half-life.

#### For alpha decay:
- Half-life \( T_{1/2} = 4 \) years
\[ \lambda_{\alpha} = \frac{\ln(2)}{4} \]

#### For beta decay:
- Half-life \( T_{1/2} = 12 \) years
\[ \lambda_{\beta} = \frac{\ln(2)}{12} \]

### Step 3: Calculate the total decay constant (λ_total)
The total decay constant when both alpha and beta decay are occurring can be calculated by adding the individual decay constants:
\[ \lambda_{total} = \lambda_{\alpha} + \lambda_{\beta} \]
Substituting the values we calculated:
\[ \lambda_{total} = \frac{\ln(2)}{4} + \frac{\ln(2)}{12} \]
To add these fractions, we need a common denominator. The least common multiple of 4 and 12 is 12.
\[ \lambda_{total} = \frac{3\ln(2)}{12} + \frac{\ln(2)}{12} = \frac{4\ln(2)}{12} = \frac{\ln(2)}{3} \]

### Step 4: Calculate the remaining percentage after 12 years
The remaining amount of the substance after time \( t \) can be calculated using the formula:
\[ N(t) = N_0 e^{-\lambda_{total} t} \]
where \( N_0 \) is the initial amount and \( t \) is the time elapsed.

Substituting \( t = 12 \) years:
\[ N(12) = N_0 e^{-\lambda_{total} \cdot 12} = N_0 e^{-\frac{\ln(2)}{3} \cdot 12} \]
This simplifies to:
\[ N(12) = N_0 e^{-4\ln(2)} = N_0 (e^{\ln(2)})^{-4} = N_0 \left(\frac{1}{2}\right)^4 = N_0 \cdot \frac{1}{16} \]

### Step 5: Calculate the percentage remaining
The percentage of the element that remains after 12 years is:
\[ \text{Percentage remaining} = \left(\frac{N(12)}{N_0}\right) \times 100 = \left(\frac{1}{16}\right) \times 100 = 6.25\% \]

### Final Answer
The percentage of the element that remains after 12 years is **6.25%**.

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