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  • Substance A has atomic mass number 16 and half-life of 1 day. Another substance has atomic mass number 32 and half life of <i

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.

Option: 1

3.38 \times 10^{24}


Option: 2

1.69 \times 10^{24}


Option: 3

6.76 \times 10^{24}


Option: 4

6.76 \times 10^{23}


Answers (1)

\left(\mathrm{N}_0\right)_{\mathrm{A}}=\frac{320}{16}=20 moles
\left(\mathrm{N}_0\right)_{\mathrm{B}}=\frac{320}{32}=10  moles
\mathrm{N}_{\mathrm{A}}=\frac{\left(\mathrm{N}_0\right)_{\mathrm{A}}}{2^{\mathrm{n}_1}}=\frac{20}{4}=5
\mathrm{N}_{\mathrm{B}}=\frac{\left(\mathrm{N}_0\right)_{\mathrm{B}}}{2^{\mathrm{n}_2}}=\frac{10}{(2)^{\frac{2}{0.5}}}=\frac{10}{2^4}=0.625
Total \mathrm{N}=5.625 moles
No. of atoms =(\mathrm{N})\left(\mathrm{N}_{\mathrm{A}}\right)

\\ =5.625 \times 6.023 \times 10^{23}=\left(3.38 \times 10^{24}\right)

Posted by

Sumit Saini

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