Question

200 Mev of energy may be obtained per fission of \[{{U}^{235}}\]. A reactor is generating 1000 kW of power, the rate of nuclear fission in the reactor is?
(A) 1000
(B) $2\times {{10}^{8}}$
(C) $3.125\times {{10}^{16}}$
(D) 931

Answer 1

Hint:We are given that 200 Mev of energy is released per fission of the nuclear reaction. The power generated by the reactor is also mentioned. Thus, we have the energy released in every fission, so we can use the concept of power to find the answer to this problem. The energy is given in Mega electron Volt while the power is given in kilo watt.

Complete step by step answer:
Energy is $200Mev$, converting into eV we get, $200\times {{10}^{6}}ev$. But the standard SI unit is Joules, thus converting into $200\times {{10}^{6}}\times 1.6\times {{10}^{-19}}J$. Also, the power is 1000 kW converting it into watts we get $1000\times 1000={{10}^{6}}W$
We know power is given by $P=\dfrac{nE}{t}$, putting the value we get,
$\Rightarrow P=\dfrac{nE}{t}$
$\Rightarrow \dfrac{P}{E}=\dfrac{n}{t}$
$\Rightarrow \dfrac{n}{t}=\dfrac{{{10}^{6}}}{200\times {{10}^{6}}\times 1.6\times {{10}^{-19}}}$
$\therefore \dfrac{n}{t}=3.125\times {{10}^{16}}$
Hence, the rate of nuclear fission comes out to be $3.125\times {{10}^{16}}$.

Hence, the correct option is C.

Additional Information:
The fusion process is that process in which two or more lighter nuclei combine to form a heavy nucleus and with the release of enormous amounts of energy. Hydrogen bombs work on the principle of nuclear fusion.

Note: While doing such problems we have to keep in mind the quantities to be taken in standard SI units of the system. The SI unit of energy is Joules and not electron Volts. Similarly, the SI unit of power is Watt. Also, power is defined as the rate of doing work that is, we have to divide the energy with time and the time is to be taken in seconds.