Substance `A` has atomic mass number `16` and half life of `1` day.Another substance `B` has atomic mass number `32` and half life of `(1)/(2)` day.If both `A` and `B` simultaneously start undergo radio activity at the same time with initial mass `320g` each,how many total atoms of `A` and `B` combined would be left after `2` days.

Options

A. `3.38times10^(24)`
B. `6.76times10^(24)`
C. `1.69times10^(24)`
D. `6.76times10^(23)`

Answer

To solve the problem, we need to find the total number of atoms of substances A and B remaining after 2 days. We will use the concept of half-life and the formula for radioactive decay.

### Step 1: Determine the initial number of atoms for both substances.

**Substance A:**
- Atomic mass number = 16
- Initial mass = 320 g
- Number of atoms (N_A) = (mass / atomic mass) × N_A (Avogadro's number)

\[ N_A = \frac{320 \, \text{g}}{16 \, \text{g/mol}} \times 6.022 \times 10^{23} \, \text{atoms/mol} \]

Calculating this gives:

\[ N_A = 20 \times 6.022 \times 10^{23} = 1.2044 \times 10^{25} \, \text{atoms} \]

**Substance B:**
- Atomic mass number = 32
- Initial mass = 320 g
- Number of atoms (N_B) = (mass / atomic mass) × N_A

\[ N_B = \frac{320 \, \text{g}}{32 \, \text{g/mol}} \times 6.022 \times 10^{23} \, \text{atoms/mol} \]

Calculating this gives:

\[ N_B = 10 \times 6.022 \times 10^{23} = 6.022 \times 10^{24} \, \text{atoms} \]

### Step 2: Calculate the decay constant (λ) for both substances.

The decay constant (λ) is given by the formula:

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

**For Substance A:**
- Half-life (T_1/2) = 1 day

\[ \lambda_A = \frac{\ln(2)}{1 \, \text{day}} \approx 0.693 \, \text{day}^{-1} \]

**For Substance B:**
- Half-life (T_1/2) = 0.5 day

\[ \lambda_B = \frac{\ln(2)}{0.5 \, \text{day}} \approx 1.386 \, \text{day}^{-1} \]

### Step 3: Calculate the remaining number of atoms after 2 days.

Using the formula for radioactive decay:

\[ N(t) = N_0 e^{-\lambda t} \]

**For Substance A:**

\[ N_A(2) = N_A(0) e^{-\lambda_A \cdot 2} \]
\[ N_A(2) = 1.2044 \times 10^{25} e^{-0.693 \cdot 2} \]
\[ N_A(2) = 1.2044 \times 10^{25} e^{-1.386} \approx 1.2044 \times 10^{25} \times 0.250 \approx 3.011 \times 10^{24} \, \text{atoms} \]

**For Substance B:**

\[ N_B(2) = N_B(0) e^{-\lambda_B \cdot 2} \]
\[ N_B(2) = 6.022 \times 10^{24} e^{-1.386 \cdot 2} \]
\[ N_B(2) = 6.022 \times 10^{24} e^{-2.772} \approx 6.022 \times 10^{24} \times 0.0625 \approx 3.764 \times 10^{23} \, \text{atoms} \]

### Step 4: Calculate the total number of remaining atoms.

\[ N_{total} = N_A(2) + N_B(2) \]
\[ N_{total} = 3.011 \times 10^{24} + 3.764 \times 10^{23} \approx 3.387 \times 10^{24} \, \text{atoms} \]

### Final Answer:
The total number of atoms of substances A and B combined left after 2 days is approximately \(3.387 \times 10^{24}\) atoms.

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