Question

The useful power produced by a reactor of $ 40\% $ efficiency , in which $ {10^{14}} $ fissions occurring each second and energy per fission in $ 275MeV $ , is
(A) $ 0.22Kw $
(B) $ 0.44kW $
(C) $ 0.88kW $
(D) $ 1.76kW $

Answer 1

Hint: The amount of energy divided by the amount of time it took to use the energy equals power. find the energy for $ {10^{14}} $ fission occurring from the given energy per fission. Use the relation between energy and power and find power for a reactor that has $ 40\% $ efficiency.

Complete answer:
Nuclear fission is the splitting of an atom's nucleus into two or more smaller nuclei. The fission process frequently generates gamma rays and releases a significant quantity of energy.
we have been given energy per fission in $ 275MeV $
So for $ {10^{14}} $ fissions we get energy equals to, Converting it into $ joules $
 $ \Rightarrow 275 \times {10^{14}} \times {10^6} \times 1.6 \times {10^{ - 19}}joules $
 $ \Rightarrow 4400joules $
A nuclear reactor, also known as an atomic pile, is a device that is used to start and control nuclear chain reactions or nuclear fusion processes. Nuclear reactors are used to generate electricity in nuclear power stations and for nuclear marine propulsion.
Now as the reactor has an efficiency of $ 40\% $ The efficiency is calculated by dividing the output power by the input power.
Efficiency, $ \eta = \dfrac{{{P_{out}}}}{{{P_{in}}}} = \dfrac{{40}}{{100}} $
Using the power and energy relation for the given reactor with an efficiency of $ 40\% $ .
 $ power = \dfrac{{40}}{{100}} \times 4400\dfrac{{joules}}{{\sec }} $
 $ \Rightarrow 1760Watt $
Converting $ Watt $ into $ kW $ we get,
 $ \Rightarrow 1.76kW $
Hence option D) $ 1.76kW $ is the correct option.

Note:
The use of nuclear reactions to generate energy is known as nuclear power. Nuclear reactions such as nuclear fission, nuclear decay, and nuclear fusion can all be used to generate electricity. Nuclear energy is the energy stored in an atom's nucleus (core).