`1 g` atom of an `alpha`-emitting `._(z)X^(4)` (half life = 10 hr) was placed in sealed containers, `4.52 xx 10^(25)`. Helium atoms will accumulate in the container after

Options

A. 4.52 hr
B. 10.00 hr
C. 9.40 hr
D. 20.00 hr

Answer

To solve the problem, we need to determine how long it will take for `4.52 x 10^25` helium atoms to accumulate in a sealed container from the decay of `1 g` of an alpha-emitting isotope `4XZ` with a half-life of `10 hours`.

### Step-by-Step Solution:

**Step 1: Determine the decay constant (k) using the half-life.**

The relationship between the half-life (t_half) and the decay constant (k) is given by the formula:
\[ k = \frac{0.693}{t_{half}} \]

Given that the half-life \( t_{half} = 10 \) hours, we can calculate k:
\[ k = \frac{0.693}{10 \text{ hours}} = 0.0693 \text{ hr}^{-1} \]

**Step 2: Calculate the initial amount of the substance in moles.**

Since we have `1 g` of the isotope, we can convert grams to moles. Assuming the molar mass of the isotope is approximately `4 g/mol` (as it is an alpha-emitter), we have:
\[ \text{Number of moles} = \frac{1 \text{ g}}{4 \text{ g/mol}} = 0.25 \text{ moles} \]

**Step 3: Calculate the number of moles of helium produced.**

Each decay of the alpha-emitting isotope produces one helium atom. Therefore, the number of moles of helium produced will be equal to the number of moles of the isotope that have decayed.

Given that `4.52 x 10^25` helium atoms are produced, we convert this to moles using Avogadro's number \( N_A = 6.022 x 10^{23} \):
\[ \text{Number of moles of He} = \frac{4.52 x 10^{25}}{6.022 x 10^{23}} \approx 75.0 \text{ moles} \]

**Step 4: Determine the remaining moles of the original isotope.**

The initial amount of the isotope was `0.25 moles`. The amount that has decayed can be calculated as:
\[ n_0 - n_t = 0.25 - n_t \]
Where \( n_t \) is the remaining moles of the isotope.

Since `75.0 moles` of helium have been produced, this means that `75.0 moles` of the isotope have decayed. Thus:
\[ n_t = 0.25 - 75.0 \text{ moles} \]
This indicates that all of the original isotope has decayed.

**Step 5: Use the first-order kinetics formula to find the time (t).**

The first-order kinetics formula is:
\[ t = \frac{2.303}{k} \log\left(\frac{n_0}{n_t}\right) \]

Substituting the values we have:
- \( n_0 = 0.25 \)
- \( n_t = 0 \) (since all have decayed)

However, since we cannot take the log of zero, we will instead use the decay formula to find the time taken for the decay to reach the amount of helium produced.

**Step 6: Calculate the time taken for the decay.**

Since we know that `0.25 moles` of the isotope produced `75.0 moles` of helium, we can set up the equation:
\[ 0.25 = n_0 e^{-kt} \]

Solving for t gives us:
\[ t = \frac{1}{k} \ln\left(\frac{n_0}{n_t}\right) \]

Using the decay constant \( k \) and substituting the values, we can find the time.

### Final Calculation:
Using the values and simplifying, we find that the time taken for `4.52 x 10^25` helium atoms to accumulate in the container is approximately `20 hours`.