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 $$\frac{1}{2}$$ day. If both $$A$$ and $$B$$ simultaneously start undergo radio activity at the same time with initial mass $$320$$ g each, how many total atoms of $$A$$ and $$B$$ combined would be left after $$2$$ days

We need to find the total number of atoms of A and B combined after 2 days.

The number of atoms remaining after time $$t$$ is: $$N = N_0 \left(\frac{1}{2}\right)^{t/t_{1/2}}$$

For substance A:

Atomic mass number = 16, half-life = 1 day, initial mass = 320 g.

Initial atoms: $$N_{0A} = \frac{320}{16} \times 6.02 \times 10^{23} = 20 \times 6.02 \times 10^{23} = 1.204 \times 10^{25}$$

After 2 half-lives: $$N_A = 1.204 \times 10^{25} \times \frac{1}{4} = 3.01 \times 10^{24}$$

For substance B:

Atomic mass number = 32, half-life = 0.5 day, initial mass = 320 g.

Initial atoms: $$N_{0B} = \frac{320}{32} \times 6.02 \times 10^{23} = 10 \times 6.02 \times 10^{23} = 6.02 \times 10^{24}$$

After 4 half-lives: $$N_B = 6.02 \times 10^{24} \times \frac{1}{16} = 3.7625 \times 10^{23}$$

Total atoms remaining:

$$N_{total} = 3.01 \times 10^{24} + 0.376 \times 10^{24} = 3.386 \times 10^{24} \approx 3.38 \times 10^{24}$$

The correct answer is Option 1: $$3.38 \times 10^{24}$$.