A `Bi^(210)` radionuclide decays via the chain `Bi^(210)(beta^(-)-"decay")/(lambda_(1))Po^(210)(alpha-decay)/(lambda_(2)) Pb^(206)` (stable), where the decay constants are `lambda_(1) = 1.6 xx 10^(-6)s^(-1), T_(1//2) ~~ 5 days , lambda_(2) = = 5.8 xx 10^(-8)s^(-1), T_(1//2) ~~ 4.6` months. `alpha & beta` activites of the `1.00 mg` a month after its manufacture is `(x)/(5) xx 10^(11)` Find `x.2^((1)/(4.6)) = 0.86`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Calculate the initial number of Bismuth-210 atoms (N₀)
Given:
- Mass of Bi-210 = 1.00 mg = 1.00 x 10^(-3) g
- Molar mass of Bi-210 = 210 g/mol
- Avogadro's number (N_A) = 6.022 x 10^(23) atoms/mol

The number of moles (n) of Bi-210 can be calculated as:
\[ n = \frac{\text{mass}}{\text{molar mass}} = \frac{1.00 \times 10^{-3} \text{ g}}{210 \text{ g/mol}} = 4.76 \times 10^{-6} \text{ mol} \]

Now, calculate the initial number of atoms (N₀):
\[ N₀ = n \times N_A = 4.76 \times 10^{-6} \text{ mol} \times 6.022 \times 10^{23} \text{ atoms/mol} \approx 2.86 \times 10^{18} \text{ atoms} \]

### Step 2: Calculate the decay of Bi-210 after 1 month
The decay constant (λ₁) for Bi-210 is given as:
\[ \lambda_1 = 1.6 \times 10^{-6} \text{s}^{-1} \]

Convert 1 month to seconds:
\[ 1 \text{ month} = 30 \text{ days} \times 24 \text{ hours/day} \times 3600 \text{ seconds/hour} = 2.628 \times 10^{6} \text{ seconds} \]

Now, calculate the number of Bi-210 atoms remaining after 1 month (N_t):
\[ N_t = N₀ e^{-\lambda_1 t} = 2.86 \times 10^{18} e^{-1.6 \times 10^{-6} \times 2.628 \times 10^{6}} \]

Calculating the exponent:
\[ -\lambda_1 t = -1.6 \times 10^{-6} \times 2.628 \times 10^{6} \approx -4.20 \]

Now, calculating N_t:
\[ N_t \approx 2.86 \times 10^{18} e^{-4.20} \approx 2.86 \times 10^{18} \times 0.015 \approx 4.29 \times 10^{16} \text{ atoms} \]

### Step 3: Calculate the beta activity (A₁)
The beta activity (A₁) is given by:
\[ A_1 = \lambda_1 N_t = 1.6 \times 10^{-6} \text{s}^{-1} \times 4.29 \times 10^{16} \text{ atoms} \approx 0.6864 \times 10^{11} \text{ disintegrations/second} \]

### Step 4: Calculate the decay of Po-210 after 1 month
The decay constant (λ₂) for Po-210 is given as:
\[ \lambda_2 = 5.8 \times 10^{-8} \text{s}^{-1} \]

Now, calculate the number of Po-210 atoms produced from the decay of Bi-210:
\[ N_{Po} = N₀ - N_t = 2.86 \times 10^{18} - 4.29 \times 10^{16} \approx 2.82 \times 10^{18} \text{ atoms} \]

Now, calculate the number of Po-210 atoms remaining after 1 month:
\[ N_{Po,t} = N_{Po} e^{-\lambda_2 t} = 2.82 \times 10^{18} e^{-5.8 \times 10^{-8} \times 2.628 \times 10^{6}} \]

Calculating the exponent:
\[ -\lambda_2 t = -5.8 \times 10^{-8} \times 2.628 \times 10^{6} \approx -0.152 \]

Now, calculating N_{Po,t}:
\[ N_{Po,t} \approx 2.82 \times 10^{18} e^{-0.152} \approx 2.82 \times 10^{18} \times 0.858 \approx 2.42 \times 10^{18} \text{ atoms} \]

### Step 5: Calculate the alpha activity (A₂)
The alpha activity (A₂) is given by:
\[ A_2 = \lambda_2 N_{Po,t} = 5.8 \times 10^{-8} \text{s}^{-1} \times 2.42 \times 10^{18} \text{ atoms} \approx 1.40 \times 10^{11} \text{ disintegrations/second} \]

### Step 6: Total activity
The total activity (A_total) is:
\[ A_{total} = A_1 + A_2 \approx 0.6864 \times 10^{11} + 1.40 \times 10^{11} \approx 2.0864 \times 10^{11} \text{ disintegrations/second} \]

### Step 7: Find x
We know that:
\[ A_{total} = \frac{x}{5} \times 10^{11} \]
Setting this equal to our total activity:
\[ 2.0864 \times 10^{11} = \frac{x}{5} \times 10^{11} \]

Solving for x:
\[ x = 2.0864 \times 5 \approx 10.432 \approx 10 \]

### Final Answer
Thus, the value of x is approximately **10**.

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