A) Zero
B) 0.125 kg
C) 0.5 kg
D) 0.25 kg
Correct Answer: B
Solution :
From Rutherford and Soddy law for radioactive decay, if N be the number of atoms of radioactive substance lift at some instant of time, then \[N={{N}_{0}}{{\left( \frac{1}{2} \right)}^{n}}\] When\[{{N}_{0}}\]is original number of atoms and\[n\]is number of half views Given, \[{{T}_{1}}/2=1600yr,{{N}_{0}}=1\,kg\] \[n=\frac{4800}{1600}=3\] \[N={{\left( \frac{1}{2} \right)}^{3}}\times 1\] \[=\frac{1}{8}=0.125\,kg\]
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