`""_(92)^(238)U` is known to undergo radioactive decay to form `""_(82)^(206)Ph` emitting alpha and beta particles. A rock initially contained `68xx10^(-6)g` of `""_(92)^(238)U`. If the number of alpha particles that it would emit during its radioactive decay of `""_(92)^(238)U` to `""_(82)^(206)Pb` in three half - lives is `Zxx10^(18)`, then what is value of Z?

Explanation

To solve the problem, we need to calculate the number of alpha particles emitted during the radioactive decay of Uranium-238 to Lead-206 over three half-lives. Let's break down the solution step by step.

### Step 1: Understand the Decay Process
Uranium-238 (\( _{92}^{238}U \)) decays to Lead-206 (\( _{82}^{206}Pb \)) by emitting alpha particles and beta particles. Each alpha decay reduces the atomic mass by 4 and the atomic number by 2.

### Step 2: Determine the Number of Alpha Particles Emitted
From the decay process:
- The decay can be represented as:
\[ _{92}^{238}U \rightarrow _{82}^{206}Pb + n \cdot _{2}^{4}He + m \cdot _{0}^{-1}e \]
- Balancing the reaction:
- Mass number: \( 238 = 206 + 4n \)
- Atomic number: \( 92 = 82 + 2n \)

From the atomic number equation:
\[ 92 - 82 = 2n \implies n = 5 \]
From the mass number equation:
\[ 238 - 206 = 4n \implies n = 8 \]
Thus, \( n = 8 \) alpha particles are emitted.

### Step 3: Calculate the Initial Amount of Uranium
The initial mass of Uranium is given as:
\[ 68 \times 10^{-6} \text{ g} \]
To find the number of moles of Uranium:
\[ \text{Moles of } U = \frac{\text{mass}}{\text{molar mass}} = \frac{68 \times 10^{-6}}{238} \approx 2.857 \times 10^{-7} \text{ moles} \]

### Step 4: Convert Moles to Atoms
Using Avogadro's number (\( 6.022 \times 10^{23} \) atoms/mole):
\[ \text{Number of atoms} = 2.857 \times 10^{-7} \text{ moles} \times 6.022 \times 10^{23} \text{ atoms/mole} \approx 1.72 \times 10^{17} \text{ atoms} \]

### Step 5: Calculate the Number of Atoms Remaining After Three Half-Lives
The formula for the remaining amount after \( n \) half-lives is:
\[ N = N_0 \left( \frac{1}{2} \right)^n \]
For three half-lives:
\[ N = 1.72 \times 10^{17} \left( \frac{1}{2} \right)^3 = 1.72 \times 10^{17} \times \frac{1}{8} = 0.215 \times 10^{17} \text{ atoms} \]

### Step 6: Calculate the Number of Atoms Decayed
The number of atoms that have decayed is:
\[ \text{Decayed atoms} = N_0 - N = 1.72 \times 10^{17} - 0.215 \times 10^{17} = 1.505 \times 10^{17} \text{ atoms} \]

### Step 7: Calculate the Total Number of Alpha Particles Emitted
Since each decay of Uranium-238 emits 8 alpha particles:
\[ \text{Total alpha particles} = 8 \times 1.505 \times 10^{17} = 1.204 \times 10^{18} \]

### Step 8: Determine the Value of Z
Given that the number of alpha particles emitted is \( Z \times 10^{18} \):
\[ Z = 1.204 \]

### Final Answer
The value of \( Z \) is approximately \( 1.2 \).