A) 996J
B) 831J
C) 499J
D) 374J
Correct Answer: B
Solution :
\[{{\operatorname{t}}_{{}^{1}/{}_{2}}} =30min\]\[{{\operatorname{N}}_{1}}={{N}_{o}}{{\left( \frac{1}{2} \right)}^{{}^{{{T}_{1}}}/{}_{30}}}{{\operatorname{N}}_{2}}={{N}_{o}}{{\left( \frac{1}{2} \right)}^{{}^{{{T}_{2}}}/{}_{30}}}\] \[{{\operatorname{N}}_{1}}=100-40=60\] \[{{\operatorname{N}}_{1}}= Atom remains undecay\] \[{{\operatorname{N}}_{2}}=100-85=15\] \[{{\operatorname{N}}_{2}}= Atom remains undecay\] \[60={{\operatorname{N}}_{o}}{{\left( \frac{1}{2} \right)}^{{}^{{{T}_{1}}}/{}_{30}}}\] ?.(1) \[15={{\operatorname{N}}_{o}}{{\left( \frac{1}{2} \right)}^{{}^{{{T}_{1}}}/{}_{30}}}\] ?.(2) By (1) and (2) \[4={{\left( \frac{1}{2} \right)}^{\frac{{{T}_{1}}-{{T}_{2}}}{30}}}\] \[4={{\left( 2 \right)}^{\frac{{{T}_{1}}-{{T}_{2}}}{30}}}\] \[{{\left( 2 \right)}^{2}}={{\left( 2 \right)}^{{}^{30}/{}_{{{T}_{1}}-{{T}_{2}}}}}\] \[2=\frac{30}{{{T}_{1}}-{{T}_{2}}}\] \[{{T}_{1}}-{{T}_{2}}=15\]
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