The half-life of `_^238U_92` against alpha decay is `4.5xx10^9` year. How much disintegration per second occurs in 1 g of `_^238U_92` ?

Explanation

To find the disintegration per second (activity) of \( _{92}^{238}U \) in 1 gram, we can follow these steps:

### Step 1: Calculate the Decay Constant (\( \lambda \))
The relationship between half-life (\( t_{1/2} \)) and the decay constant (\( \lambda \)) is given by the formula:

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

Given that the half-life of \( _{92}^{238}U \) is \( 4.5 \times 10^9 \) years, we first convert this to seconds:

\[ t_{1/2} = 4.5 \times 10^9 \text{ years} \times 3.154 \times 10^7 \text{ seconds/year} \approx 1.419 \times 10^{17} \text{ seconds} \]

Now, substituting this value into the decay constant formula:

\[ \lambda = \frac{\ln(2)}{1.419 \times 10^{17}} \approx \frac{0.693}{1.419 \times 10^{17}} \approx 4.88 \times 10^{-18} \text{ s}^{-1} \]

### Step 2: Calculate the Number of Nuclei (\( N \)) in 1 gram of \( _{92}^{238}U \)
The number of nuclei can be calculated using Avogadro's number (\( N_A \)) and the molar mass of \( _{92}^{238}U \):

\[ N_A \approx 6.022 \times 10^{23} \text{ nuclei/mole} \]

The molar mass of \( _{92}^{238}U \) is approximately 238 g/mole. Therefore, in 1 gram of \( _{92}^{238}U \):

\[ N = \frac{N_A}{238} \approx \frac{6.022 \times 10^{23}}{238} \approx 2.53 \times 10^{21} \text{ nuclei} \]

### Step 3: Calculate the Activity (\( A \))
The activity \( A \) (disintegration per second) can be calculated using the formula:

\[ A = \lambda N \]

Substituting the values we have:

\[ A = (4.88 \times 10^{-18} \text{ s}^{-1})(2.53 \times 10^{21} \text{ nuclei}) \approx 1.237 \times 10^4 \text{ disintegrations/second} \]

### Final Answer
The disintegration per second occurring in 1 g of \( _{92}^{238}U \) is approximately:

\[ A \approx 1.237 \times 10^4 \text{ disintegrations/second} \]

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