A nuclear reaction of $$40\%$$ efficiency has $$1014$$ decays / second. If the energy obtained per fission is $$250 \,MeV,$$ then power of reactor is:
Correct option is C. $$1.6 \,kW$$
Per fission received energy $$-250\ MeV$$
$$=250 \times 10^6 \times 1.6 \times 10^{-19}\ J$$
Number of fissions per second $$=10^{14}$$
$$\therefore$$ Per second received energy
$$=250 \times 10^6 \times 1.6 \times 10^{-19}\ \times 10^{14}$$
As efficiency is $$40\%$$, so output power
$$P=250 \times 10^6 \times 1.6 \times 10^{-5} \times \dfrac{40}{100}$$
$$=1.6 \times 10^3 \ watt=1.6\ kW$$