The half life of `._92^238U` undergoing `alpha`-decay is `4.5xx10^9` years. The activity of 1 g sample of `._92^238U` is

Options

A. `1.23xx10^4` Bq
B. `1.23xx10^5` Bq
C. `1.23xx10^3` Bq
D. `1.23xx10^6` Bq

Explanation

To solve the problem of finding the activity of a 1 g sample of \( _{92}^{238}U \) undergoing alpha decay, we will follow these steps:

### Step 1: Convert Half-Life to Seconds
The half-life of \( _{92}^{238}U \) is given as \( 4.5 \times 10^9 \) years. We need to convert this into seconds.

\[ \text{Half-life in seconds} = 4.5 \times 10^9 \text{ years} \times 365 \text{ days/year} \times 24 \text{ hours/day} \times 3600 \text{ seconds/hour} \]

Calculating this gives:

\[ \text{Half-life in seconds} = 4.5 \times 10^9 \times 365 \times 24 \times 3600 \approx 1.42 \times 10^{17} \text{ seconds} \]

### Step 2: Calculate the Decay Constant (\( \lambda \))
The decay constant \( \lambda \) can be calculated using the formula:

\[ \lambda = \frac{\ln(2)}{T_{1/2}} \]

Where \( T_{1/2} \) is the half-life in seconds. The value of \( \ln(2) \) is approximately \( 0.693 \).

Substituting the values:

\[ \lambda = \frac{0.693}{1.42 \times 10^{17}} \approx 4.87 \times 10^{-18} \text{ s}^{-1} \]

### Step 3: Calculate the Number of Atoms in 1 g of \( _{92}^{238}U \)
To find the number of atoms, we first need to calculate the number of moles in 1 g of \( _{92}^{238}U \):

\[ \text{Number of moles} = \frac{\text{mass}}{\text{molar mass}} = \frac{1 \text{ g}}{238 \text{ g/mol}} \approx 0.00420 \text{ mol} \]

Now, using Avogadro's number (\( 6.022 \times 10^{23} \) atoms/mol), we can find the total number of atoms:

\[ N = \text{Number of moles} \times \text{Avogadro's number} = 0.00420 \times 6.022 \times 10^{23} \approx 2.53 \times 10^{21} \text{ atoms} \]

### Step 4: Calculate the Activity (\( A \))
The activity \( A \) can be calculated using the formula:

\[ A = \lambda N \]

Substituting the values we have:

\[ A = (4.87 \times 10^{-18} \text{ s}^{-1}) \times (2.53 \times 10^{21} \text{ atoms}) \approx 1.23 \times 10^4 \text{ disintegrations per second} \]

### Final Answer
The activity of a 1 g sample of \( _{92}^{238}U \) is approximately \( 1.23 \times 10^4 \) Bq (where 1 Bq = 1 disintegration per second).

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