Two radioactive elements R and S disintegrate as
`RtoP+alpha,lamda_(R)=4.5xx10^(-3)"years"^(-1)`
`StoQ+betal,lamda_(S)=3xx10^(-3)"years"^(-1)`
Starting with number of atoms of R and S in the ratio of 2:1 this ratio after the lapse of three half lives of R will be

Options

A. `3:2`
B. `1:3`
C. `1:1`
D. `2:1`

Explanation

To solve the problem, we need to find the ratio of the number of atoms of radioactive elements R and S after 3 half-lives of R. Let's go through the solution step by step.

### Step 1: Determine the half-lives of R and S
The half-life (T₁/₂) of a radioactive substance is related to its decay constant (λ) by the formula:
\[ T_{1/2} = \frac{\ln(2)}{\lambda} \]

For element R:
\[ \lambda_R = 4.5 \times 10^{-3} \text{ years}^{-1} \]
\[ T_{1/2(R)} = \frac{\ln(2)}{4.5 \times 10^{-3}} \approx 154.3 \text{ years} \]

For element S:
\[ \lambda_S = 3 \times 10^{-3} \text{ years}^{-1} \]
\[ T_{1/2(S)} = \frac{\ln(2)}{3 \times 10^{-3}} \approx 231.0 \text{ years} \]

### Step 2: Calculate the number of half-lives
We are interested in the situation after 3 half-lives of R. We need to find out how many half-lives of S correspond to 3 half-lives of R.

Using the ratio of half-lives:
\[ \text{Ratio of half-lives} = \frac{T_{1/2(S)}}{T_{1/2(R)}} = \frac{231.0}{154.3} \approx 1.5 \]
This means that 3 half-lives of R correspond to:
\[ \text{Number of half-lives of S} = 3 \times \frac{154.3}{231.0} \approx 2 \]

### Step 3: Initial ratio of atoms
Initially, the number of atoms of R and S is in the ratio of 2:1. Let’s denote:
- Initial number of atoms of R = 2N
- Initial number of atoms of S = N

### Step 4: Calculate remaining atoms after decay
The remaining number of atoms after n half-lives can be calculated using the formula:
\[ N_t = N_0 \left(\frac{1}{2}\right)^n \]

For R after 3 half-lives:
\[ N_R = 2N \left(\frac{1}{2}\right)^3 = 2N \times \frac{1}{8} = \frac{N}{4} \]

For S after 2 half-lives:
\[ N_S = N \left(\frac{1}{2}\right)^2 = N \times \frac{1}{4} = \frac{N}{4} \]

### Step 5: Calculate the final ratio
Now we can find the ratio of the remaining atoms of R to S:
\[ \text{Final ratio} = \frac{N_R}{N_S} = \frac{\frac{N}{4}}{\frac{N}{4}} = 1:1 \]

### Conclusion
After 3 half-lives of R, the ratio of the number of atoms of R to S will be 1:1.