A) 1:9:15
B) 2:3:5
C) 5:3:2
D) 125:30:8
Correct Answer: D
Solution :
As we know that \[{{T}_{1/2}}=4.47\times {{10}^{8}}yr\] So \[{{R}_{1}}:{{R}_{2}}:{{R}_{3}}=\frac{{{l}_{1}}}{{{A}_{1}}}:\frac{{{l}_{2}}}{{{A}_{2}}}:\frac{{{l}_{3}}}{{{A}_{3}}}\] \[=\frac{l_{1}^{2}}{{{V}_{1}}}:\frac{l_{2}^{2}}{{{V}_{2}}}:\frac{l_{3}^{2}}{{{V}_{3}}}\] \[=\frac{l_{1}^{2}}{({{m}_{1}}d)}:\frac{l_{2}^{2}}{({{m}_{2}}d)}:\frac{l_{3}^{2}}{({{m}_{3}}d)}\] \[=\frac{l_{1}^{2}}{{{m}_{1}}}:\frac{l_{2}^{2}}{{{m}_{2}}}:\frac{l_{3}^{2}}{{{m}_{3}}}\] \[n\times 0.3=1-0.778=0.22\] \[n=\frac{0.222}{0.3}=0.74\]
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