Find the half life of `U^(238)`, if one gram of it emits `1.24xx10^4` `alpha`-particle per second. Avogadro's Number `=6.023xx10^(23)`.

Options

A. a The half life of this nuclide is `4.5xx10^(9)` years
B. b The half-life of this nuclide is `9xx10^(9)` years
C. c The activity of the prepared smaple is `2.48xx10^(4)` particles/sec
D. d The acitivty of the prpared sample is `1.24xx10^(4)` particles/sec

Answer

To find the half-life of Uranium-238, we can use the relationship between the activity of a radioactive substance, the decay constant, and the number of atoms present. Here's a step-by-step solution:

### Step 1: Understand the given data
- Activity (A) = 1.24 × 10^4 alpha particles per second
- Avogadro's Number (N_A) = 6.023 × 10^23 atoms/mol
- Molar mass of Uranium-238 (M) = 238 g/mol

### Step 2: Calculate the number of atoms in 1 gram of Uranium-238
The number of moles of Uranium-238 in 1 gram can be calculated using:
\[ \text{Number of moles} = \frac{\text{mass}}{\text{molar mass}} = \frac{1 \text{ g}}{238 \text{ g/mol}} = \frac{1}{238} \text{ mol} \]
Now, we can find the number of atoms (n) in 1 gram:
\[ n = \text{Number of moles} \times N_A = \frac{1}{238} \times 6.023 \times 10^{23} \approx 2.53 \times 10^{21} \text{ atoms} \]

### Step 3: Relate activity to decay constant
The activity (A) is related to the decay constant (λ) and the number of atoms (n) by the formula:
\[ A = \lambda n \]
Rearranging this gives us:
\[ \lambda = \frac{A}{n} \]

### Step 4: Substitute the values to find λ
Substituting the values we have:
\[ \lambda = \frac{1.24 \times 10^4}{2.53 \times 10^{21}} \approx 4.90 \times 10^{-18} \text{ s}^{-1} \]

### Step 5: Relate decay constant to half-life
The relationship between the decay constant (λ) and the half-life (t_half) is given by:
\[ t_{half} = \frac{\ln(2)}{\lambda} \]
Substituting the value of λ:
\[ t_{half} = \frac{\ln(2)}{4.90 \times 10^{-18}} \approx \frac{0.693}{4.90 \times 10^{-18}} \approx 1.41 \times 10^{17} \text{ s} \]

### Step 6: Convert seconds to years
To convert seconds to years, we use the conversion factor:
\[ 1 \text{ year} \approx 3.15 \times 10^7 \text{ seconds} \]
Thus,
\[ t_{half} \approx \frac{1.41 \times 10^{17}}{3.15 \times 10^7} \approx 4.48 \times 10^9 \text{ years} \]

### Final Answer
The half-life of Uranium-238 is approximately \(4.5 \times 10^9\) years.

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