A fresh radioactive mixture containing short lived species X and Y. Both the species together emitting 8000 `alpha` -particles per minute initially. 20 minutes later X was emitting `alpha` - particles at the rate of 4500 per minute. If the half lives of the species X and Y are 10 minute and 500 hours, then what is the ratio of initial activities of X and Y in the mixture

Explanation

To solve the problem, we need to find the ratio of the initial activities of the radioactive species X and Y in the mixture. Let's break down the steps.

### Step 1: Define Initial Activities
Let the initial activities of species X and Y be \( A_X^0 \) and \( A_Y^0 \) respectively. According to the problem, the total initial activity is:
\[ A_X^0 + A_Y^0 = 8000 \text{ particles/minute} \]

### Step 2: Determine Activity of X After 20 Minutes
The half-life of species X is given as 10 minutes. After 20 minutes, which is two half-lives, the activity of X will be reduced to:
\[ A_X = A_X^0 \times \left(\frac{1}{2}\right)^{\frac{20}{10}} = A_X^0 \times \left(\frac{1}{2}\right)^2 = A_X^0 \times \frac{1}{4} \]

### Step 3: Set Up the Equation for Total Activity After 20 Minutes
After 20 minutes, the activity of X is \( \frac{A_X^0}{4} \). The activity of Y, which has a half-life of 500 hours (very long compared to 20 minutes), will remain approximately the same as its initial activity \( A_Y^0 \). Therefore, the total activity after 20 minutes is:
\[ A_X + A_Y \approx \frac{A_X^0}{4} + A_Y^0 = 3500 \text{ particles/minute} \]

### Step 4: Substitute Initial Activity of Y
From the first equation, we can express \( A_Y^0 \) in terms of \( A_X^0 \):
\[ A_Y^0 = 8000 - A_X^0 \]
Substituting this into the total activity equation gives:
\[ \frac{A_X^0}{4} + (8000 - A_X^0) = 3500 \]

### Step 5: Solve for \( A_X^0 \)
Now, simplify the equation:
\[ \frac{A_X^0}{4} + 8000 - A_X^0 = 3500 \]
Combining like terms:
\[ 8000 - \frac{3A_X^0}{4} = 3500 \]
Subtract 8000 from both sides:
\[ -\frac{3A_X^0}{4} = 3500 - 8000 \]
\[ -\frac{3A_X^0}{4} = -4500 \]
Multiplying both sides by -1:
\[ \frac{3A_X^0}{4} = 4500 \]
Now, multiply both sides by \( \frac{4}{3} \):
\[ A_X^0 = 4500 \times \frac{4}{3} = 6000 \]

### Step 6: Find \( A_Y^0 \)
Substituting \( A_X^0 \) back into the equation for \( A_Y^0 \):
\[ A_Y^0 = 8000 - 6000 = 2000 \]

### Step 7: Calculate the Ratio
Now that we have both initial activities:
\[ \text{Ratio of } A_X^0 \text{ to } A_Y^0 = \frac{A_X^0}{A_Y^0} = \frac{6000}{2000} = 3 \]

### Final Answer
The ratio of the initial activities of X and Y is:
\[ \boxed{3} \]