Question

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.

• A. $3.38 \times 10^{24}$
• B. $6.76 \times 10^{24}$
• C. $6.76 \times 10^{23}$
• D. $1.69 \times 10^{24}$

Answer 1

Answer: (A)

$ \left( N _0\right) A =\frac{320}{16}=20 \text { moles } $
$ \left( N _0\right) B =\frac{320}{32}=10 \text { moles } $
$ N _{ A }=\frac{\left( N _0\right)_{ A }}{(2)^{2 / 1}}=\frac{20}{4}=5 $
$N _{ B }=\frac{\left( N _0\right)_{ B }}{(2)^{2 / 5}}=\frac{10}{2^4}=0.625$
$ \text { Total } N =5.625 $
$ \text { No. of atoms }=5.625 \times 6.023 \times 10^{23}$
$=3.38 \times 10^{24}$