The decay constant for an `alpha-` decay of `Th^(232)` is `1.58xx10^(-10)s^(-1)`. How many `alpha-`decays occur from `1 g` sample in 365 days ?

Options

A. `2.89xx10^(-19)`
B. `1.298xx10^(19)`
C. `8.219xx10^(19)`
D. None of these

Explanation

To solve the problem of how many alpha decays occur from a 1 g sample of Th-232 in 365 days, we can follow these steps:

### Step 1: Calculate the total time in seconds
First, we need to convert 365 days into seconds.

\[ \text{Total time (T)} = 365 \text{ days} \times 24 \text{ hours/day} \times 60 \text{ minutes/hour} \times 60 \text{ seconds/minute} \]

Calculating this gives:

\[ T = 365 \times 24 \times 60 \times 60 = 31,536,000 \text{ seconds} \]

### Step 2: Use the decay constant to find the fraction of Th-232 remaining
The decay constant (\( \lambda \)) is given as \( 1.58 \times 10^{-10} \, \text{s}^{-1} \). We can use the formula for radioactive decay:

\[ N_t = N_0 e^{-\lambda T} \]

Where:
- \( N_t \) is the amount remaining after time \( T \)
- \( N_0 \) is the initial amount (1 g in this case)
- \( \lambda \) is the decay constant
- \( T \) is the time in seconds

Substituting the values:

\[ N_t = 1 \, \text{g} \cdot e^{- (1.58 \times 10^{-10}) \cdot (31,536,000)} \]

Calculating the exponent:

\[ \lambda T = (1.58 \times 10^{-10}) \cdot (31,536,000) \approx 4.976 \]

Now substituting back:

\[ N_t = 1 \cdot e^{-4.976} \approx 1 \cdot 0.0067 \approx 0.955 \, \text{g} \]

### Step 3: Calculate the amount of Th-232 that has decayed
The amount that has decayed (\( N_d \)) is given by:

\[ N_d = N_0 - N_t \]

Substituting the values:

\[ N_d = 1 \, \text{g} - 0.955 \, \text{g} = 0.045 \, \text{g} \]

### Step 4: Calculate the number of alpha particles emitted
To find the number of alpha particles emitted, we use Avogadro's number (\( 6.022 \times 10^{23} \)) and the molar mass of Th-232 (approximately 232 g/mol):

\[ \text{Number of moles of Th-232 decayed} = \frac{N_d}{\text{Molar mass}} = \frac{0.045 \, \text{g}}{232 \, \text{g/mol}} \approx 0.000194 \, \text{mol} \]

Now, using Avogadro's number to find the number of atoms:

\[ \text{Number of alpha particles} = \text{Number of moles} \times \text{Avogadro's number} = 0.000194 \, \text{mol} \times 6.022 \times 10^{23} \, \text{atoms/mol} \approx 1.17 \times 10^{20} \text{ alpha particles} \]

### Final Answer
The total number of alpha decays that occur from a 1 g sample of Th-232 in 365 days is approximately \( 1.17 \times 10^{20} \).

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