The activity of a sample of radioactive element `X^(100)` is 6.02 curie. Its decay constant is `3.7 xx 10^(4) s^(-1)`.
The initial mass of the sample will be
a. `10^(-6) g` b. `10^(-8) g` c. `10^(-20) g` d. `10^(-15) g`

Options

A. a. `10^(-6) g`
B. b. `10^(-8) g`
C. c. `10^(-20) g`
D. d. `10^(-15) g`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the relationship between activity, decay constant, and number of nuclei
The activity \( A \) of a radioactive sample is given by the formula:
\[ A = \lambda N \]
where:
- \( A \) is the activity,
- \( \lambda \) is the decay constant,
- \( N \) is the number of radioactive nuclei.

### Step 2: Convert the activity from curies to decays per second
Given that the activity \( A = 6.02 \) curie, we convert this to decays per second using the conversion factor:
\[ 1 \text{ curie} = 3.7 \times 10^{10} \text{ decays/second} \]
Thus,
\[ A = 6.02 \times 3.7 \times 10^{10} \text{ decays/second} \]
Calculating this gives:
\[ A = 22.294 \times 10^{10} \text{ decays/second} \approx 2.2294 \times 10^{11} \text{ decays/second} \]

### Step 3: Rearrange the activity formula to find \( N \)
We can rearrange the activity formula to solve for \( N \):
\[ N = \frac{A}{\lambda} \]
Substituting the values we have:
\[ N = \frac{2.2294 \times 10^{11}}{3.7 \times 10^{4}} \]
Calculating this gives:
\[ N \approx 6.022 \times 10^{6} \]

### Step 4: Relate the number of nuclei to the mass of the sample
The number of nuclei \( N \) can also be expressed in terms of the mass \( m \) of the sample:
\[ N = \frac{m}{M} \times N_A \]
where:
- \( m \) is the mass of the sample,
- \( M \) is the atomic mass (100 for element \( X^{100} \)),
- \( N_A \) is Avogadro's number (\( 6.022 \times 10^{23} \) nuclei/mol).

Rearranging gives:
\[ m = \frac{N \times M}{N_A} \]

### Step 5: Substitute known values to find the mass
Substituting the values we have:
\[ m = \frac{6.022 \times 10^{6} \times 100}{6.022 \times 10^{23}} \]
This simplifies to:
\[ m = \frac{6.022 \times 10^{8}}{6.022 \times 10^{23}} = 10^{-15} \text{ g} \]

### Conclusion
The initial mass of the sample is:
\[ \boxed{10^{-15} \text{ g}} \]
Thus, the correct answer is option (d).