A) \[{{\cos }^{-1}}\left( \frac{1}{2} \right)\]
B) \[{{\cos }^{-1}}\left( \frac{-1}{6} \right)\]
C) \[{{\cos }^{-1}}\left( \frac{-1}{3} \right)\]
D) \[{{\cos }^{-1}}\left( \frac{1}{4} \right)\]
Correct Answer: C
Solution :
Vector perpendicular to face \[\frac{N}{{{N}_{0}}}=\frac{60}{100}={{\left( \frac{1}{2} \right)}^{n}}\] \[\Rightarrow \] \[{{2}^{n}}=\frac{10}{6}\] \[\Rightarrow \] \[n\times 0.3=1-0.778=0.22\] \[n=\frac{0.222}{0.3}=0.74\] Vector perpendicular to face \[ABC={{n}_{2}}\] \[=AB\times AC\] \[=(-3\hat{i}+3\hat{j})\times (\hat{j}+\hat{k})\] \[=\left| \begin{matrix} {\hat{i}} & {\hat{j}} & {\hat{k}} \\ -3 & 3 & 0 \\ 0 & 1 & 1 \\ \end{matrix} \right|\]\[=3\hat{i}+3\hat{j}-3\hat{k}\] Since, angle between faces equals to angle between their normals. \[t=n{{T}_{1/2}}=0.74\times 4.47\times {{10}^{8}}\] \[t=3.3\times {{10}^{8}}yr\] \[Y\left( n,\alpha \right)\] \[\alpha -\] \[_{Z}{{Y}^{A}}{{+}_{0}}{{n}^{1}}{{\to }_{3}}L{{i}^{7}}{{+}_{2}}H{{e}^{4}}\] \[\Rightarrow \]You need to login to perform this action.
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