A) - 5.40 eV
B) - 2.72 eV
C) - 0.85 eV
D) - 0.54 eV
Correct Answer: D
Solution :
[d] \[A={{A}_{0}}{{\left( \frac{1}{2} \right)}^{\frac{t}{{{T}_{1/2}}}}}\Rightarrow \frac{1}{64}={{\left( \frac{1}{2} \right)}^{\frac{60}{{{T}_{1/2}}}}}\] \[\Rightarrow \]\[{{E}_{5}}=\frac{-13.6}{{{5}^{2}}}=\frac{-13.6}{25}=-0.54\ eV\] |
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