A nuclear power plant supplying electrical power to a village uses a radioactive material of half life T years as the fuel. The amount of fuel at the beginning is such that the total power requirement of the village is $$12.5$$% of the electrical power available from the plant at the time. If the plant is able to meet the total power needs of the village for a maximum period of nT years, then the value of n is

Question

A nuclear power plant supplying electrical power to a village uses a radioactive material of half life T years as the fuel. The amount of fuel at the beginning is such that the total power requirement of the village is $$12.5$$% of the electrical power available from the plant at the time. If the plant is able to meet the total power needs of the village for a maximum period of  nT years, then the value of  n is

Infinity Learn · Accepted Answer

Pfinal $$=12.5100Pinitial$$ ​⇒$$PfinalPinitial=18=NE1N0E1=12n$$​$$E1=Energy$$ emitted per $$sec$$​⇒$$n=3$$,no of half lives

Radioactivity - Nuclei

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$\begin{array}{l} P_{final} = \frac{12.5}{100} P_{initial} \\ \Rightarrow \frac{P_{f i n a l}}{P_{i n i t i a l}} = \frac{1}{8} = \frac{(N) E^{1}}{N_{0} E^{1}} = \frac{1}{2^{n}} \\ E^{1} = E n e r g y e m i t t e d p e r s e c \\ \Rightarrow n = 3, n o o f h a l f l i v e s \end{array}$

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Radioactivity - Nuclei

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Pfinal =12.5100Pinitial ​⇒PfinalPinitial=18=NE1N0E1=12n​E1=Energy emitted per sec​⇒n=3,no of half lives

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$\begin{array}{l} P_{final} = \frac{12.5}{100} P_{initial} \\ \Rightarrow \frac{P_{f i n a l}}{P_{i n i t i a l}} = \frac{1}{8} = \frac{(N) E^{1}}{N_{0} E^{1}} = \frac{1}{2^{n}} \\ E^{1} = E n e r g y e m i t t e d p e r s e c \\ \Rightarrow n = 3, n o o f h a l f l i v e s \end{array}$